RS Aggarwal Solutions for Class 11 Chapter 23 Ellipse

Question 1. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \)
Answer: Given: \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \)

Since 25 > 9, the equation takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Comparing, we get \( a^2 = 25 \) and \( b^2 = 9 \), so \( a = 5 \) and \( b = 3 \).

(i) Length of Major Axes: Since \( a > b \), the major axis lies along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times 5 = 10 \text{ units} \)

(ii) Coordinates of the Vertices: Since \( a > b \), the vertex coordinates are \( (a, 0) \) and \( (-a, 0) \).
\( \text{Vertices} = (5, 0) \text{ and } (-5, 0) \)

(iii) Coordinates of the Foci: The foci are given by \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \).
\( c^2 = 25 - 9 = 16 \)
\( c = 4 \)
\( \text{Foci} = (\pm 4, 0) \)

(iv) Eccentricity: \( e = \frac{c}{a} = \frac{4}{5} \)

(v) Length of the Latus Rectum: \( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 9}{5} = \frac{18}{5} \)

Exam Tip: Always check which denominator is larger to identify the major axis orientation. The relationship \( c^2 = a^2 - b^2 \) must be used consistently to find foci coordinates.

Question 2. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( \frac{x^2}{49} + \frac{y^2}{36} = 1 \)
Answer: Given: \( \frac{x^2}{49} + \frac{y^2}{36} = 1 \)

Since 49 > 36, the equation is of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Comparing, we obtain \( a^2 = 49 \) and \( b^2 = 36 \), giving \( a = 7 \) and \( b = 6 \).

(i) Length of Major Axes: Since \( a > b \), the major axis is along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times 7 = 14 \text{ units} \)

(ii) Coordinates of the Vertices: Since \( a > b \), vertices occur at \( (a, 0) \) and \( (-a, 0) \).
\( \text{Vertices} = (7, 0) \text{ and } (-7, 0) \)

(iii) Coordinates of the Foci: Foci are at \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \).
\( c^2 = 49 - 36 = 13 \)
\( c = \sqrt{13} \)
\( \text{Foci} = (\pm\sqrt{13}, 0) \)

(iv) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{13}}{7} \)

(v) Length of the Latus Rectum: \( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 36}{7} = \frac{72}{7} \)

Exam Tip: When both denominators are perfect squares, extract the roots directly. Remember that eccentricity is always between 0 and 1 for an ellipse.

Question 3. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 16x^2 + 25y^2 = 400 \)
Answer: Given: \( 16x^2 + 25y^2 = 400 \)

Dividing both sides by 400, we obtain:
\( \frac{16}{400}x^2 + \frac{25}{400}y^2 = 1 \)
\( \frac{x^2}{25} + \frac{y^2}{16} = 1 \)

Since 25 > 16, the equation is of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Comparing, we get \( a^2 = 25 \) and \( b^2 = 16 \), so \( a = 5 \) and \( b = 4 \).

(i) Length of Major Axes: Since \( a > b \), the major axis runs along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times 5 = 10 \text{ units} \)

(ii) Coordinates of the Vertices: Since \( a > b \), vertices are positioned at \( (a, 0) \) and \( (-a, 0) \).
\( \text{Vertices} = (5, 0) \text{ and } (-5, 0) \)

(iii) Coordinates of the Foci: Foci are located at \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \).
\( c^2 = 25 - 16 = 9 \)
\( c = 3 \)
\( \text{Foci} = (\pm 3, 0) \)

(iv) Eccentricity: \( e = \frac{c}{a} = \frac{3}{5} \)

(v) Length of the Latus Rectum: \( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 16}{5} = \frac{32}{5} \)

Exam Tip: Always convert non-standard form equations to standard form first by dividing through. This step prevents arithmetic errors in identifying a and b values.

Question 4. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( x^2 + 4y^2 = 100 \)
Answer: Given: \( x^2 + 4y^2 = 100 \)

Dividing both sides by 100, we get:
\( \frac{x^2}{100} + \frac{4y^2}{100} = 1 \)
\( \frac{x^2}{100} + \frac{y^2}{25} = 1 \)

Since 100 > 25, the equation takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Comparing, we obtain \( a^2 = 100 \) and \( b^2 = 25 \), giving \( a = 10 \) and \( b = 5 \).

(i) Length of Major Axes: Since \( a > b \), the major axis extends along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times 10 = 20 \text{ units} \)

(ii) Coordinates of the Vertices: Since \( a > b \), vertex coordinates are \( (a, 0) \) and \( (-a, 0) \).
\( \text{Vertices} = (10, 0) \text{ and } (-10, 0) \)

(iii) Coordinates of the Foci: Foci are positioned at \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \).
\( c^2 = 100 - 25 = 75 \)
\( c = \sqrt{75} = 5\sqrt{3} \)
\( \text{Foci} = (\pm 5\sqrt{3}, 0) \)

(iv) Eccentricity: \( e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} \)

(v) Length of the Latus Rectum: \( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 25}{10} = 5 \)

Exam Tip: Simplify square roots where possible, such as \( \sqrt{75} = 5\sqrt{3} \). This makes calculations cleaner and shows mathematical maturity in handling radicals.

Question 5. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 9x^2 + 16y^2 = 144 \)
Answer: Given: \( 9x^2 + 16y^2 = 144 \)

Dividing both sides by 144, we obtain:
\( \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \)
\( \frac{x^2}{16} + \frac{y^2}{9} = 1 \)

Since 16 > 9, the equation is of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Comparing, we get \( a^2 = 16 \) and \( b^2 = 9 \), so \( a = 4 \) and \( b = 3 \).

(i) Length of Major Axes: Since \( a > b \), the major axis lies along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times 4 = 8 \text{ units} \)

(ii) Coordinates of the Vertices: Since \( a > b \), vertices are at \( (a, 0) \) and \( (-a, 0) \).
\( \text{Vertices} = (4, 0) \text{ and } (-4, 0) \)

(iii) Coordinates of the Foci: Foci are at \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \).
\( c^2 = 16 - 9 = 7 \)
\( c = \sqrt{7} \)
\( \text{Foci} = (\pm\sqrt{7}, 0) \)

(iv) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{7}}{4} \)

(v) Length of the Latus Rectum: \( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2} \)

Exam Tip: When the coefficient of \( x^2 \) is smaller than the coefficient of \( y^2 \) in the original equation, the major axis is horizontal. Always verify this relationship before proceeding with calculations.

Question 6. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 4x^2 + 9y^2 = 1 \)
Answer: Given: \( 4x^2 + 9y^2 = 1 \)

Rewriting in standard form:
\( \frac{4x^2}{1} + \frac{9y^2}{1} = 1 \)
\( \frac{x^2}{1/4} + \frac{y^2}{1/9} = 1 \)

Since \( \frac{1}{4} > \frac{1}{9} \), the equation is of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Comparing, we get \( a^2 = \frac{1}{4} \) and \( b^2 = \frac{1}{9} \), so \( a = \frac{1}{2} \) and \( b = \frac{1}{3} \).

(i) Length of Major Axes: Since \( a > b \), the major axis extends along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times \frac{1}{2} = 1 \text{ unit} \)

(ii) Coordinates of the Vertices: Since \( a > b \), vertices occur at \( (a, 0) \) and \( (-a, 0) \).
\( \text{Vertices} = \left(\frac{1}{2}, 0\right) \text{ and } \left(-\frac{1}{2}, 0\right) \)

(iii) Coordinates of the Foci: Foci are positioned at \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \).
\( c^2 = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \)
\( c = \frac{\sqrt{5}}{6} \)
\( \text{Foci} = \left(\pm\frac{\sqrt{5}}{6}, 0\right) \)

(iv) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{5}/6}{1/2} = \frac{\sqrt{5}}{6} \times 2 = \frac{\sqrt{5}}{3} \)

(v) Length of the Latus Rectum: \( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times (1/9)}{1/2} = \frac{2}{9} \times 2 = \frac{4}{9} \)

Exam Tip: When working with fractional values of a and b, handle arithmetic carefully, especially when computing differences and ratios. Double-check your simplifications to avoid careless errors.

Question 1. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 4x^2 + 9y^2 = 1 \)
Answer: Given: \( 4x^2 + 9y^2 = 1 \)

Rewriting in standard form: \( \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{9}} = 1 \)

Since \( \frac{1}{4} > \frac{1}{9} \), we have \( a^2 = \frac{1}{4} \) and \( b^2 = \frac{1}{9} \)

Therefore \( a = \frac{1}{2} \) and \( b = \frac{1}{3} \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the x-axis.
\( \text{Length of major axes} = 2a = 2 \times \frac{1}{2} = 1 \) unit

(ii) Coordinates of the vertices:
\( \text{Vertices} = \left(\frac{1}{2}, 0\right) \text{ and } \left(-\frac{1}{2}, 0\right) \)

(iii) Coordinates of the foci:
\( c^2 = a^2 - b^2 = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36} \)

\( c = \frac{\sqrt{5}}{6} \)

\( \text{Coordinates of foci} = \left(\pm\frac{\sqrt{5}}{6}, 0\right) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\frac{\sqrt{5}}{6}}{\frac{1}{2}} = \frac{\sqrt{5}}{6} \times 2 = \frac{\sqrt{5}}{3} \)

(v) Length of the latus rectum:
\( \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \times \left(\frac{1}{3}\right)^2}{\frac{1}{2}} = \frac{\frac{2}{9}}{\frac{1}{2}} = \frac{2}{9} \times 2 = \frac{4}{9} \)

Exam Tip: Always check which denominator is larger to determine the major axis direction. For the latus rectum, use the formula \( \frac{2b^2}{a} \) accurately with all fractional values.

Question 2. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \)
Answer: Given: \( \frac{x^2}{4} + \frac{y^2}{25} = 1 \)

Since \( 4 < 25 \), we have \( a^2 = 25 \) and \( b^2 = 4 \)

Therefore \( a = 5 \) and \( b = 2 \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 5 = 10 \) units

(ii) Coordinates of the vertices:
\( \text{Vertices} = (0, 5) \text{ and } (0, -5) \)

(iii) Coordinates of the foci:
\( c^2 = a^2 - b^2 = 25 - 4 = 21 \)

\( c = \sqrt{21} \)

\( \text{Coordinates of foci} = (0, \pm\sqrt{21}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{21}}{5} \)

(v) Length of the latus rectum:
\( \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \times (2)^2}{5} = \frac{8}{5} \)

Exam Tip: When the y-denominator is larger, the major axis is vertical. Always compute \( c^2 = a^2 - b^2 \) carefully before finding the foci coordinates.

Question 3. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \)
Answer: Given: \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \)

Since \( 9 < 16 \), we have \( a^2 = 16 \) and \( b^2 = 9 \)

Therefore \( a = 4 \) and \( b = 3 \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 4 = 8 \) units

(ii) Coordinates of the vertices:
\( \text{Vertices} = (0, 4) \text{ and } (0, -4) \)

(iii) Coordinates of the foci:
\( c^2 = a^2 - b^2 = 16 - 9 = 7 \)

\( c = \sqrt{7} \)

\( \text{Coordinates of foci} = (0, \pm\sqrt{7}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{7}}{4} \)

(v) Length of the latus rectum:
\( \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \times (3)^2}{4} = \frac{18}{4} = \frac{9}{2} \)

Exam Tip: Compare both denominators carefully. The larger denominator gives \( a^2 \), determining the major axis direction and ensuring all five properties are calculated in the correct order.

Question 4. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 3x^2 + 2y^2 = 18 \)
Answer: Given: \( 3x^2 + 2y^2 = 18 \)

Dividing both sides by 18: \( \frac{x^2}{6} + \frac{y^2}{9} = 1 \)

Since \( 6 < 9 \), we have \( a^2 = 9 \) and \( b^2 = 6 \)

Therefore \( a = 3 \) and \( b = \sqrt{6} \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 3 = 6 \) units

(ii) Coordinates of the vertices:
\( \text{Vertices} = (0, 3) \text{ and } (0, -3) \)

(iii) Coordinates of the foci:
\( c^2 = a^2 - b^2 = 9 - 6 = 3 \)

\( c = \sqrt{3} \)

\( \text{Coordinates of foci} = (0, \pm\sqrt{3}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{3}}{3} \)

(v) Length of the latus rectum:
\( \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \times (\sqrt{6})^2}{3} = \frac{2 \times 6}{3} = \frac{12}{3} = 4 \)

Exam Tip: Always convert the ellipse to standard form by dividing through by the constant. When \( b \) contains a square root, simplify \( b^2 \) before computing the latus rectum.

Question 5. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 9x^2 + y^2 = 36 \)
Answer: Given: \( 9x^2 + y^2 = 36 \)

Dividing both sides by 36: \( \frac{x^2}{4} + \frac{y^2}{36} = 1 \)

Since \( 4 < 36 \), we have \( a^2 = 36 \) and \( b^2 = 4 \)

Therefore \( a = 6 \) and \( b = 2 \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 6 = 12 \) units

(ii) Coordinates of the vertices:
\( \text{Vertices} = (0, 6) \text{ and } (0, -6) \)

(iii) Coordinates of the foci:
\( c^2 = a^2 - b^2 = 36 - 4 = 32 \)

\( c = \sqrt{32} = 4\sqrt{2} \)

\( \text{Coordinates of foci} = (0, \pm\sqrt{32}) = (0, \pm 4\sqrt{2}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{32}}{6} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \)

(v) Length of the latus rectum:
\( \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \times (2)^2}{6} = \frac{8}{6} = \frac{4}{3} \)

Exam Tip: Simplify radicals like \( \sqrt{32} \) to \( 4\sqrt{2} \) for cleaner final answers. Always double-check the comparison of denominators to identify the major axis correctly.

Question 6. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. \( 16x^2 + y^2 = 16 \)
Answer: Given: \( 16x^2 + y^2 = 16 \)

Dividing both sides by 16: \( \frac{x^2}{1} + \frac{y^2}{16} = 1 \)

Since \( 1 < 16 \), we have \( a^2 = 16 \) and \( b^2 = 1 \)

Therefore \( a = 4 \) and \( b = 1 \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 4 = 8 \) units

(ii) Coordinates of the vertices:
\( \text{Vertices} = (0, 4) \text{ and } (0, -4) \)

(iii) Coordinates of the foci:
\( c^2 = a^2 - b^2 = 16 - 1 = 15 \)

\( c = \sqrt{15} \)

\( \text{Coordinates of foci} = (0, \pm\sqrt{15}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{15}}{4} \)

(v) Length of the latus rectum:
\( \text{Length of latus rectum} = \frac{2b^2}{a} = \frac{2 \times (1)^2}{4} = \frac{2}{4} = \frac{1}{2} \)

Exam Tip: Even when one denominator equals 1, follow the standard comparison process. The latus rectum formula remains the same regardless of the size of the semi-minor axis.

Question 11. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. 16x² + y² = 16
Answer: Given: \( 16x^2 + y^2 = 16 \)

Divide both sides by 16:
\( x^2 + \frac{y^2}{16} = 1 \) …(i)

Since 1 < 16, the equation is in the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) …(ii)

Comparing equations (i) and (ii):
\( a^2 = 16 \) and \( b^2 = 1 \)
\( \implies a = 4 \) and \( b = 1 \)

(i) Length of major axes: Since \( a > b \), the major axis lies along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 4 = 8 \) units

(ii) Coordinates of the Vertices:
\( \text{Vertices} = (0, \pm a) = (0, \pm 4) \)

(iii) Coordinates of the foci: We know that \( c^2 = a^2 - b^2 \)
\( c^2 = 16 - 1 = 15 \)
\( c = \sqrt{15} \)
\( \text{Coordinates of foci} = (0, \pm\sqrt{15}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{15}}{4} \)

(v) Length of the Latus Rectum:
\( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 1}{4} = \frac{1}{2} \)
In simple words: Transform the given equation into standard form by dividing both sides by 16. Identify which parameter is larger to determine the orientation of the major axis. Apply the formulas for each property - vertices are at the endpoints of the major axis, foci depend on the value c found from the relationship \( c^2 = a^2 - b^2 \), eccentricity measures how elongated the ellipse is, and the latus rectum is a specific chord through each focus.

Exam Tip: Always convert the equation to standard form first. Remember that the larger denominator indicates the major axis direction, and use the relationship \( c^2 = a^2 - b^2 \) consistently for all calculations.

Question 12. Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. 25x² + 4y² = 100
Answer: Given: \( 25x^2 + 4y^2 = 100 \)

Divide both sides by 100:
\( \frac{x^2}{4} + \frac{y^2}{25} = 1 \) …(i)

Since 4 < 25, the equation is in the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) …(ii)

Comparing equations (i) and (ii):
\( a^2 = 25 \) and \( b^2 = 4 \)
\( \implies a = 5 \) and \( b = 2 \)

(i) Length of major axes: Since \( a > b \), the major axis is along the y-axis.
\( \text{Length of major axes} = 2a = 2 \times 5 = 10 \) units

(ii) Coordinates of the Vertices:
\( \text{Vertices} = (0, \pm a) = (0, \pm 5) \)

(iii) Coordinates of the foci: We know that \( c^2 = a^2 - b^2 \)
\( c^2 = 25 - 4 = 21 \)
\( c = \sqrt{21} \)
\( \text{Coordinates of foci} = (0, \pm\sqrt{21}) \)

(iv) Eccentricity:
\( e = \frac{c}{a} = \frac{\sqrt{21}}{5} \)

(v) Length of the Latus Rectum:
\( \text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5} \)
In simple words: Start by converting the equation to standard form by dividing both sides by 100. The larger denominator (25 versus 4) tells you the major axis is vertical. Use standard formulas to find vertices on the major axis, calculate the focal distance using \( c^2 = a^2 - b^2 \), find eccentricity as the ratio \( c/a \), and compute the latus rectum using \( 2b^2/a \).

Exam Tip: Ensure you correctly identify which denominator is larger to determine axis orientation. Double-check your arithmetic when computing \( c^2 \) and simplify all final answers completely.

Question 13. Find the equation of the ellipse whose vertices are at (±6, 0) and foci at (±4, 0).
Answer: Given: Vertices \( = (\pm 6, 0) \) …(i)

Since vertices are of the form \( (\pm a, 0) \) …(ii), the major axis lies along the x-axis.

From equations (i) and (ii):
\( a = 6 \implies a^2 = 36 \)

The standard form of the ellipse with major axis along the x-axis is:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Given: Coordinates of foci \( = (\pm 4, 0) \) …(iii)

Since foci are of the form \( (\pm c, 0) \) …(iv):
\( c = 4 \)

We know that \( c^2 = a^2 - b^2 \):
\( 16 = 36 - b^2 \)
\( b^2 = 20 \)

Substituting \( a^2 = 36 \) and \( b^2 = 20 \) into the standard form:
\( \frac{x^2}{36} + \frac{y^2}{20} = 1 \)
In simple words: The vertices give you the value of a (the semi-major axis length), which is 6. The foci provide c (the distance from center to each focus), which is 4. Using the relationship \( c^2 = a^2 - b^2 \), you solve for b to find the semi-minor axis. Then substitute both values into the standard ellipse equation.

Exam Tip: Identify the major axis direction from the vertex coordinates - if vertices are on the x-axis, use \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Always verify that \( a > b \) in your final answer.

Question 14. Find the equation of the ellipse whose vertices are at (0, ±4) and foci at (0, ±√7).
Answer: Given: Vertices \( = (0, \pm 4) \) …(i)

Since vertices are of the form \( (0, \pm a) \) …(ii), the major axis lies along the y-axis.

From equations (i) and (ii):
\( a = 4 \implies a^2 = 16 \)

The standard form of the ellipse with major axis along the y-axis is:
\( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \)

Given: Coordinates of foci \( = (0, \pm\sqrt{7}) \) …(iii)

Since foci are of the form \( (0, \pm c) \) …(iv):
\( c = \sqrt{7} \)

We know that \( c^2 = a^2 - b^2 \):
\( 7 = 16 - b^2 \)
\( b^2 = 9 \)

Substituting \( a^2 = 16 \) and \( b^2 = 9 \) into the standard form:
\( \frac{x^2}{9} + \frac{y^2}{16} = 1 \)
In simple words: Since the vertices lie on the y-axis at (0, ±4), you know a = 4 and the major axis is vertical. The foci at (0, ±√7) give you c = √7. Apply the formula \( c^2 = a^2 - b^2 \) to find b² = 9, then write the equation using the form where y² has the larger denominator.

Exam Tip: Watch the position of vertices to determine which term (x² or y²) has the larger denominator. For vertical major axes, y² comes with the larger denominator a².

Question 15. Find the equation of the ellipse the ends of whose major and minor axes are (±4, 0) and (0, ±3) respectively.
Answer: Given:
Ends of Major Axis \( = (\pm 4, 0) \)
Ends of Minor Axis \( = (0, \pm 3) \)

Since the major axis lies along the x-axis, the standard form is:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) …(i)

where a is the semi-major axis and b is the semi-minor axis.

From the given endpoints:
\( a = 4 \) and \( b = 3 \)

Substituting into equation (i):
\( \frac{x^2}{(4)^2} + \frac{y^2}{(3)^2} = 1 \)

\( \frac{x^2}{16} + \frac{y^2}{9} = 1 \)
In simple words: The endpoints of the major axis show how far the ellipse extends horizontally, giving you a = 4. The endpoints of the minor axis show how far it extends vertically, giving you b = 3. Substitute these into the standard form where x² is divided by a² and y² is divided by b².

Exam Tip: The longer distance between endpoints tells you which is the major axis. Always ensure the larger value goes with the larger denominator in your equation.

Question 16. The length of the major axis of an ellipse is 20 units, and its foci are (±5√3, 0). Find the equation of the ellipse.
Answer: Let the equation of the ellipse be:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Given: Length of Major Axis \( = 20 \) units …(i)

We know that Length of Major Axis \( = 2a \) …(ii)

From equations (i) and (ii):
\( 2a = 20 \implies a = 10 \)

Given: Coordinates of foci \( = (\pm 5\sqrt{3}, 0) \) …(iii)

We know that Coordinates of foci \( = (\pm c, 0) \) …(iv)

From equations (iii) and (iv):
\( c = 5\sqrt{3} \)

We know that \( c^2 = a^2 - b^2 \):
\( (5\sqrt{3})^2 = (10)^2 - b^2 \)
\( 75 = 100 - b^2 \)
\( b^2 = 25 \)

Substituting \( a^2 = 100 \) and \( b^2 = 25 \) into the standard form:
\( \frac{x^2}{100} + \frac{y^2}{25} = 1 \)
In simple words: The major axis length gives you 2a = 20, so a = 10. The foci at \( (\pm 5\sqrt{3}, 0) \) tell you c = 5√3. Use \( c^2 = a^2 - b^2 \) to find b² = 25. Plug these values into the standard equation form.

Exam Tip: Remember to square values when using \( c^2 = a^2 - b^2 \). Double-check that \( a^2 > b^2 \) in your final equation.

Question 17. Find the equation of the ellipse whose foci are (±2, 0) and the eccentricity is 1/2.
Answer: Let the equation of the ellipse be:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Given: Coordinates of foci \( = (\pm 2, 0) \) …(i)

We know that Coordinates of foci \( = (\pm c, 0) \) …(ii)

From equations (i) and (ii):
\( c = 2 \)

Given: Eccentricity \( = \frac{1}{2} \)

We know that \( e = \frac{c}{a} \):
\( \frac{1}{2} = \frac{2}{a} \)
\( a = 4 \)

We know that \( c^2 = a^2 - b^2 \):
\( (2)^2 = (4)^2 - b^2 \)
\( 4 = 16 - b^2 \)
\( b^2 = 12 \)

Substituting \( a^2 = 16 \) and \( b^2 = 12 \) into the standard form:
\( \frac{x^2}{16} + \frac{y^2}{12} = 1 \)
In simple words: The focal coordinates (±2, 0) give c = 2. The eccentricity e = 1/2 is the ratio c/a, so you can find a = 4. Then use \( c^2 = a^2 - b^2 \) to calculate b² = 12, and write the equation.

Exam Tip: Eccentricity is always less than 1 for an ellipse. Use the relationship e = c/a to link focal distance with the semi-major axis.

Question 18. Find the equation of the ellipse whose foci are at (±1, 0) and eccentricity is 1/2.
Answer: Let the equation of the ellipse be:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Given: Coordinates of foci \( = (\pm 1, 0) \) …(i)

We know that Coordinates of foci \( = (\pm c, 0) \) …(ii)

From equations (i) and (ii):
\( c = 1 \)

Given: Eccentricity \( = \frac{1}{2} \)

We know that \( e = \frac{c}{a} \):
\( \frac{1}{2} = \frac{1}{a} \)
\( a = 2 \)

We know that \( c^2 = a^2 - b^2 \):
\( (1)^2 = (2)^2 - b^2 \)
\( 1 = 4 - b^2 \)
\( b^2 = 3 \)

Substituting \( a^2 = 4 \) and \( b^2 = 3 \) into the standard form:
\( \frac{x^2}{4} + \frac{y^2}{3} = 1 \)
In simple words: The foci at (±1, 0) indicate c = 1 and the major axis is horizontal. Using eccentricity e = c/a = 1/2, you get a = 2. Apply \( c^2 = a^2 - b^2 \) to find b² = 3, then substitute into the standard ellipse equation.

Exam Tip: When foci are on the x-axis, the equation has a² under x² and b² under y². Verify your answer by checking that the calculated b² is positive and less than a².

Question 19. Find the equation of the ellipse whose foci are at (0, ±4) and eccentricity = 4/5.
Answer: The foci are located at (0, ±4), which indicates the major and minor axes lie on the y and x axes respectively. From the foci positions, c = 4. Using the eccentricity formula e = c/a, we get 4/5 = 4/a, so a = 5. Applying the relationship c² = a² - b², we find (4)² = (5)² - b², giving 16 = 25 - b², hence b² = 9. Since the foci of the ellipse rest on the y - axis, the equation takes the form x²/b² + y²/a² = 1. Substituting the calculated values yields the final equation:

\( \frac{x^2}{9} + \frac{y^2}{25} = 1 \)
In simple words: When the foci sit on the vertical axis, divide x² by the smaller number and y² by the larger number, then set the sum equal to 1.

Exam Tip: Always identify which axis contains the foci to determine the correct ellipse form. The larger denominator should correspond to the axis with the foci.

Question 20. Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points (4, 3) and (-1, 4).
Answer: The center sits at the origin with the major axis running along the x - axis, so the equation has the form x²/a² + y²/b² = 1. Since the curve passes through (4, 3), substituting gives 16/a² + 9/b² = 1. For the point (-1, 4), substituting yields 1/a² + 16/b² = 1. Multiplying the second equation by 16 produces 16/a² + 256/b² = 16. Subtracting this from the first equation results in 9/b² - 256/b² = 1 - 16, which simplifies to -247/b² = -15, giving b² = 247/15. Substituting b² back into the second equation: 1/a² + 16/(247/15) = 1, which leads to 1/a² + 240/247 = 1. Solving for a² yields (247 - 240)/247 = 7/247, so a² = 247/7. Substituting both values into the standard form:

\( \frac{7x^2 + 15y^2}{247} = 1 \)

or \( 7x^2 + 15y^2 = 247 \)
In simple words: Replace x and y with the given points' coordinates to get two equations. Solve this pair by elimination to find a² and b², then build the final ellipse equation.

Exam Tip: Always verify your answer by checking that both given points satisfy the final equation.

Question 21. Find the equation of the ellipse with eccentricity 3/4, foci on the y-axis, center at the origin and passing through the point (6, 4).
Answer: With eccentricity e = 3/4, the relationship e = c/a gives 3/4 = c/a, so c = 3a/4. Using c² = a² - b², we get (3a/4)² = a² - b², which simplifies to 9a²/16 = a² - b². Rearranging yields b² = a² - 9a²/16 = 7a²/16. Since the foci lie on the y - axis, the equation has the form x²/b² + y²/a² = 1. Substituting b² = 7a²/16 gives x²/(7a²/16) + y²/a² = 1, or 16x²/7a² + y²/a² = 1. The ellipse passes through (6, 4), so 16(36)/7a² + 16/a² = 1. Simplifying: 576/7a² + 16/a² = 1, which gives (576 + 112)/7a² = 1, hence 688/7a² = 1, yielding a² = 688/7. Then b² = 7 × (688/7) / 16 = 688/16. Substituting both values:

\( \frac{16x^2}{688} + \frac{7y^2}{688} = 1 \)

or \( 16x^2 + 7y^2 = 688 \)
In simple words: Use the eccentricity to link a and b. Then use the given point to find a². Finally, calculate b² and write the ellipse equation.

Exam Tip: When foci are on the y - axis, remember that a > b and the a² term appears under y², not x².

Question 22. Find the equation of the ellipse which passes through the point (4, 1) and having its foci at (±3, 0).
Answer: Let the ellipse equation be x²/a² + y²/b² = 1. The foci sit at (±3, 0), so c = 3. Using c² = a² - b², we get 9 = a² - b², hence b² = a² - 9. The ellipse passes through (4, 1), so 16/a² + 1/b² = 1. Substituting b² = a² - 9 yields 16/a² + 1/(a² - 9) = 1. Multiplying through by a²(a² - 9) gives 16(a² - 9) + a² = a²(a² - 9), which simplifies to 16a² - 144 + a² = a⁴ - 9a². Rearranging: a⁴ - 26a² + 144 = 0. Factoring yields (a² - 8)(a² - 18) = 0. If a² = 8, then b² = 8 - 9 = -1, which is impossible since b² must be positive. If a² = 18, then b² = 18 - 9 = 9, which is valid. Therefore, the equation is:

\( \frac{x^2}{18} + \frac{y^2}{9} = 1 \)
In simple words: Use the foci to link a and b. Substitute the given point to create an equation in a². Solve for a², reject any negative value for b², and write the final equation.

Exam Tip: Always check that calculated values produce a valid ellipse (b² must remain positive). Reject solutions that violate this condition.

Question 23. Find the equation of an ellipse, the lengths of whose major and minor axes are 10 and 8 units respectively.
Answer: The length of the major axis equals 2a = 10, giving a = 5. The length of the minor axis equals 2b = 8, yielding b = 4. Assuming the major axis lies on the x - axis and the center sits at the origin, the standard form is x²/a² + y²/b² = 1. Substituting the calculated values produces:

\( \frac{x^2}{25} + \frac{y^2}{16} = 1 \)
In simple words: Divide each axis length by 2 to get a and b. Then place a² under x² and b² under y² in the standard ellipse equation.

Exam Tip: Remember that the major axis length is always 2a and the minor axis length is always 2b, regardless of axis orientation.

Question 24. Find the equation of an ellipse whose eccentricity is 2/3, the latus rectum is 5, and the center is at the origin.
Answer: Let the ellipse equation be x²/a² + y²/b² = 1. Given eccentricity e = 2/3, we have e = c/a, so 2/3 = c/a, which yields c = 2a/3. Using c² = a² - b², we obtain (2a/3)² = a² - b², giving 4a²/9 = a² - b². Solving for b²: b² = a² - 4a²/9 = 5a²/9. The latus rectum is given by 2b²/a = 5. Substituting b² = 5a²/9 yields 2(5a²/9)/a = 5, which simplifies to 10a/9 = 5, so a = 9/2. Then b² = 5(9/2)²/9 = 5(81/4)/9 = 405/36 = 45/4. Therefore, the equation is:

\( \frac{x^2}{81/4} + \frac{y^2}{45/4} = 1 \)

Multiplying by 4: \( \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \)

or in simplified form: \( 20x^2 + 36y^2 = 405 \)
In simple words: Use eccentricity to relate a and b. Use the latus rectum formula to find a. Then calculate b² and substitute into the standard equation.

Exam Tip: The latus rectum formula 2b²/a is a key relationship - memorize it and apply it carefully with the given numerical value.

Question 25. Find the eccentricity of an ellipse whose latus rectum is one half of its minor axis.
Answer: Let the equation of the needed ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) ...(i) We are told that the length of the latus rectum equals half the length of the minor axis. Since the length of the latus rectum is \( \frac{2b^2}{a} \) and the length of the minor axis is 2b, we can write: \[ \frac{2b^2}{a} = \frac{1}{2} \times 2b \] \[ \frac{2b^2}{a} = b \] \[ 2b = a \quad \text{...(ii)} \] To find the eccentricity, we use the relationship \( e = \frac{c}{a} \) ...(iii), where \( c^2 = a^2 - b^2 \). Substituting \( a = 2b \) from equation (ii): \[ c^2 = (2b)^2 - b^2 \] \[ c^2 = 4b^2 - b^2 = 3b^2 \] \[ c = b\sqrt{3} \] Substituting the values of c and a into equation (iii): \[ e = \frac{c}{a} = \frac{b\sqrt{3}}{2b} = \frac{\sqrt{3}}{2} \]
In simple words: When the latus rectum (a chord through the focus perpendicular to the major axis) is half the minor axis, you can set up an equation linking these measurements. This leads to a relationship between a and b, which lets you calculate the eccentricity as \( \frac{\sqrt{3}}{2} \).

Exam Tip: Remember the formula for latus rectum length: \( \frac{2b^2}{a} \). Always express the given condition as an equation and solve for the relationship between a and b before finding eccentricity.

Question 26. Find the eccentricity of an ellipse whose latus rectum is one half of its major axis.
Answer: Let the equation of the needed ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) ...(i) We are told that the length of the latus rectum equals one-half the length of the major axis. Since the length of the latus rectum is \( \frac{2b^2}{a} \) and the length of the major axis is 2a, we can write: \[ \frac{2b^2}{a} = \frac{1}{2} \times 2a \] \[ \frac{2b^2}{a} = a \] \[ 2b^2 = a^2 \quad \text{...(ii)} \] To find the eccentricity, we use the relationship \( e = \frac{c}{a} \) ...(iii), where \( c^2 = a^2 - b^2 \). From equation (ii), \( a^2 = 2b^2 \), so: \[ c^2 = 2b^2 - b^2 = b^2 \] \[ c = b \] Also, \( a = b\sqrt{2} \). Substituting the values of c and a into equation (iii): \[ e = \frac{c}{a} = \frac{b}{b\sqrt{2}} = \frac{1}{\sqrt{2} \]
In simple words: When the latus rectum is half the major axis, you establish a connection between a and b. This relationship shows that the eccentricity simplifies to \( \frac{1}{\sqrt{2}} \), which is smaller than in the previous problem.

Exam Tip: Compare Questions 25 and 26 to see how changing the condition (from half the minor axis to half the major axis) changes the eccentricity value. This helps you understand the geometry of ellipses better.