Ellipse JEE Mathematics Worksheets Set 01

Subjective Questions

Question. (a) Find the equation of the ellipse with its centre (1, 2), focus at (6, 2) and passing through the point (4, 6).
Answer: Let the equation of the ellipse be \( \frac{(x-1)^2}{a^2} + \frac{(y-2)^2}{b^2} = 1 \).
Centre is (1, 2) and focus is (6, 2).
The distance between centre and focus is \( ae = \sqrt{(6-1)^2 + (2-2)^2} = 5 \).
\( \implies \) \( a^2e^2 = 25 \)
\( \implies \) \( a^2 - b^2 = 25 \)
\( \implies \) \( b^2 = a^2 - 25 \).
The ellipse passes through (4, 6):
\( \frac{(4-1)^2}{a^2} + \frac{(6-2)^2}{b^2} = 1 \)
\( \implies \) \( \frac{9}{a^2} + \frac{16}{b^2} = 1 \)
Substituting \( b^2 \):
\( \frac{9}{a^2} + \frac{16}{a^2 - 25} = 1 \)
\( \implies \) \( 9(a^2 - 25) + 16a^2 = a^2(a^2 - 25) \)
\( \implies \) \( 9a^2 - 225 + 16a^2 = a^4 - 25a^2 \)
\( \implies \) \( a^4 - 50a^2 + 225 = 0 \)
\( \implies \) \( (a^2 - 45)(a^2 - 5) = 0 \).
Since \( a^2 > 25 \), we take \( a^2 = 45 \).
Then \( b^2 = 45 - 25 = 20 \).
The equation is \( \frac{(x-1)^2}{45} + \frac{(y-2)^2}{20} = 1 \).

Question. (b) An ellipse passes through the points (–3, 1) & (2, –2) & its principal axis are along the coordinate axes in order. Find its equation.
Answer: Let the equation be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Passing through (–3, 1):
\( \frac{9}{a^2} + \frac{1}{b^2} = 1 \) ....(i)
Passing through (2, –2):
\( \frac{4}{a^2} + \frac{4}{b^2} = 1 \) ....(ii)
Multiply (i) by 4:
\( \frac{36}{a^2} + \frac{4}{b^2} = 4 \) ....(iii)
Subtract (ii) from (iii):
\( \frac{32}{a^2} = 3 \)
\( \implies \) \( a^2 = \frac{32}{3} \).
Substitute \( a^2 \) in (i):
\( \frac{9}{32/3} + \frac{1}{b^2} = 1 \)
\( \implies \) \( \frac{27}{32} + \frac{1}{b^2} = 1 \)
\( \implies \) \( \frac{1}{b^2} = 1 - \frac{27}{32} = \frac{5}{32} \)
\( \implies \) \( b^2 = \frac{32}{5} \).
The equation is \( \frac{3x^2}{32} + \frac{5y^2}{32} = 1 \) or \( 3x^2 + 5y^2 = 32 \).

Question. The tangent at the point \( \alpha \) on a standard ellipse meets the auxiliary circle in two points which subtends a right angle at the centre. Show that the eccentricity of the ellipse is \( (1 + \sin^2 \alpha)^{-1/2} \).
Answer: Equation of tangent at \( \alpha \): \( \frac{x \cos \alpha}{a} + \frac{y \sin \alpha}{b} = 1 \).
Auxiliary circle: \( x^2 + y^2 = a^2 \).
Homogenizing the circle with the tangent:
\( x^2 + y^2 = a^2 \left( \frac{x \cos \alpha}{a} + \frac{y \sin \alpha}{b} \right)^2 \)
\( \implies \) \( x^2 + y^2 = a^2 \left( \frac{x^2 \cos^2 \alpha}{a^2} + \frac{y^2 \sin^2 \alpha}{b^2} + \frac{2xy \sin \alpha \cos \alpha}{ab} \right) \)
The chords subtend a right angle at the centre, so sum of coefficients of \( x^2 \) and \( y^2 \) is zero:
\( (1 - \cos^2 \alpha) + (1 - \frac{a^2 \sin^2 \alpha}{b^2}) = 0 \)
\( \implies \) \( \sin^2 \alpha + 1 = \frac{a^2 \sin^2 \alpha}{b^2} \)
\( \implies \) \( \frac{b^2}{a^2} = \frac{\sin^2 \alpha}{1 + \sin^2 \alpha} \).
Eccentricity \( e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{\sin^2 \alpha}{1 + \sin^2 \alpha} = \frac{1}{1 + \sin^2 \alpha} \)
\( \implies \) \( e = (1 + \sin^2 \alpha)^{-1/2} \).

Question. If the normal at the point P(\( \theta \)) to the ellipse \( \frac{x^2}{14} + \frac{y^2}{5} = 1 \), intersects it again at the point Q(2\( \theta \)), show that \( \cos \theta = – (2/3) \).
Answer: Equation of normal at \( \theta \): \( ax \sec \theta - by \text{ cosec } \theta = a^2 - b^2 \).
Here \( a^2 = 14, b^2 = 5 \), so \( a^2 - b^2 = 9 \).
Normal: \( \sqrt{14} x \sec \theta - \sqrt{5} y \text{ cosec } \theta = 9 \).
It passes through \( Q(2\theta) \), i.e., \( (\sqrt{14} \cos 2\theta, \sqrt{5} \sin 2\theta) \):
\( \sqrt{14} (\sqrt{14} \cos 2\theta) \sec \theta - \sqrt{5} (\sqrt{5} \sin 2\theta) \text{ cosec } \theta = 9 \)
\( \implies \) \( 14 \frac{\cos 2\theta}{\cos \theta} - 5 \frac{2 \sin \theta \cos \theta}{\sin \theta} = 9 \)
\( \implies \) \( 14 \frac{2 \cos^2 \theta - 1}{\cos \theta} - 10 \cos \theta = 9 \)
\( \implies \) \( 28 \cos^2 \theta - 14 - 10 \cos^2 \theta = 9 \cos \theta \)
\( \implies \) \( 18 \cos^2 \theta - 9 \cos \theta - 14 = 0 \).
Solving the quadratic equation in \( \cos \theta \):
\( \cos \theta = \frac{9 \pm \sqrt{81 - 4(18)(-14)}}{2(18)} = \frac{9 \pm \sqrt{81 + 1008}}{36} = \frac{9 \pm 33}{36} \).
\( \cos \theta = \frac{42}{36} = \frac{7}{6} \) (not possible) or \( \cos \theta = \frac{-24}{36} = -\frac{2}{3} \).
So \( \cos \theta = -2/3 \).

Question. If the normals at the points P, Q, R with eccentric angles \( \alpha, \beta, \gamma \) on the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) are concurrent, then show that, \( \begin{vmatrix} \sin \alpha & \cos \alpha & \sin 2\alpha \\ \sin \beta & \cos \beta & \sin 2\beta \\ \sin \gamma & \cos \gamma & \sin 2\gamma \end{vmatrix} = 0 \).
Answer: Equation of normal at any point \( \theta \): \( (a \sec \theta)x - (b \text{ cosec } \theta)y = a^2 - b^2 \).
If normals at \( \alpha, \beta, \gamma \) are concurrent at \( (h, k) \):
\( (a \sec \alpha)h - (b \text{ cosec } \alpha)k = a^2 - b^2 \)
\( \implies \) \( ah \cos \alpha - bk \sin \alpha = (a^2 - b^2) \sin \alpha \cos \alpha \)
\( \implies \) \( 2ah \cos \alpha - 2bk \sin \alpha = (a^2 - b^2) \sin 2\alpha \).
Similarly for \( \beta \) and \( \gamma \). For these three equations to have a non-trivial solution in \( (2ah, -2bk, a^2-b^2) \):
\( \begin{vmatrix} \cos \alpha & \sin \alpha & \sin 2\alpha \\ \cos \beta & \sin \beta & \sin 2\beta \\ \cos \gamma & \sin \gamma & \sin 2\gamma \end{vmatrix} = 0 \).
Rearranging the columns gives the desired result: \( \begin{vmatrix} \sin \alpha & \cos \alpha & \sin 2\alpha \\ \sin \beta & \cos \beta & \sin 2\beta \\ \sin \gamma & \cos \gamma & \sin 2\gamma \end{vmatrix} = 0 \).

Question. Prove that the equation to the circle, having double contact with the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (with eccentricity e) at the ends of a latus rectum, is \( x^2 + y^2 – 2ae^3x = a^2 (1 – e^2 – e^4) \).
Answer: Let the circle be \( (x - h)^2 + y^2 = r^2 \).
Double contact at \( x = ae \), where \( y^2 = b^2(1 - e^2) \).
At point of contact: \( (ae - h)^2 + b^2(1 - e^2) = r^2 \) ....(i)
Also, the slopes of the tangent to circle and ellipse must be equal at contact point.
Differentiating ellipse: \( \frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y' = -\frac{b^2x}{a^2y} \).
Differentiating circle: \( 2(x - h) + 2yy' = 0 \implies y' = -\frac{x - h}{y} \).
Setting them equal at \( x = ae \):
\( \frac{b^2(ae)}{a^2y} = \frac{ae - h}{y} \)
\( \implies \) \( \frac{b^2e}{a} = ae - h \)
\( \implies \) \( h = ae - \frac{a(1 - e^2)e}{a} \cdot a \) (since \( b^2 = a^2(1 - e^2) \))
\( \implies \) \( h = ae - ae(1 - e^2) = ae^3 \).
From (i): \( r^2 = (ae - ae^3)^2 + a^2(1 - e^2)(1 - e^2) \)
\( \implies \) \( r^2 = a^2e^2(1 - e^2)^2 + a^2(1 - e^2)^2 = a^2(1 - e^2)^2 (e^2 + 1) \)
\( \implies \) \( r^2 = a^2(1 - 2e^2 + e^4)(1 + e^2) = a^2(1 - e^2 - e^4 + e^6) \).
Equation: \( (x - ae^3)^2 + y^2 = a^2(1 - e^2 - e^4 + e^6) \)
\( \implies \) \( x^2 - 2ae^3x + a^2e^6 + y^2 = a^2 - a^2e^2 - a^2e^4 + a^2e^6 \)
\( \implies \) \( x^2 + y^2 - 2ae^3x = a^2(1 - e^2 - e^4) \).

Question. Find the equations of the lines with equal intercepts on the axes & which touch the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).
Answer: Equal intercepts means the line is of the form \( \frac{x}{c} + \frac{y}{c} = 1 \) or \( \frac{x}{c} - \frac{y}{c} = 1 \).
This implies the slope \( m = \pm 1 \).
Condition for tangency to \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( c^2 = a^2m^2 + b^2 \).
Here \( a^2 = 16, b^2 = 9 \), and \( m^2 = 1 \).
\( c^2 = 16(1) + 9 = 25 \)
\( \implies \) \( c = \pm 5 \).
The equations of the tangents are \( y = \pm x \pm 5 \).
Specifically, for equal intercepts: \( x + y = \pm 5 \) and \( x - y = \pm 5 \).

Question. Suppose x and y are real numbers and that \( x^2 + 9y^2 – 4x + 6y + 4 = 0 \) then find the maximum value of (4x – 9y).
Answer: The given equation is \( (x - 2)^2 + 9(y + \frac{1}{3})^2 - 4 - 1 + 4 = 0 \)
\( \implies \) \( (x - 2)^2 + 9(y + \frac{1}{3})^2 = 1 \)
\( \implies \) \( (x - 2)^2 + \frac{(y + 1/3)^2}{(1/3)^2} = 1 \).
Let \( x - 2 = \cos \theta \) and \( y + \frac{1}{3} = \frac{1}{3} \sin \theta \).
\( \implies \) \( x = 2 + \cos \theta \) and \( y = -\frac{1}{3} + \frac{1}{3} \sin \theta \).
We need to find maximum value of \( Z = 4x - 9y \):
\( Z = 4(2 + \cos \theta) - 9(-\frac{1}{3} + \frac{1}{3} \sin \theta) \)
\( Z = 8 + 4 \cos \theta + 3 - 3 \sin \theta = 11 + 4 \cos \theta - 3 \sin \theta \).
Maximum value of \( 4 \cos \theta - 3 \sin \theta \) is \( \sqrt{4^2 + (-3)^2} = 5 \).
So, \( Z_{max} = 11 + 5 = 16 \).

Question. A tangent having slope – 4/3 to the ellipse \( \frac{x^2}{18} + \frac{y^2}{32} = 1 \), intersects the axis of x & y in points A & B respectively. If O is the origin, find the area of triangle OAB.
Answer: Equation of tangent with slope \( m = -4/3 \):
\( y = mx \pm \sqrt{a^2m^2 + b^2} \).
Here \( a^2 = 18, b^2 = 32, m = -4/3 \).
\( a^2m^2 + b^2 = 18(\frac{16}{9}) + 32 = 32 + 32 = 64 \).
Tangent: \( y = -\frac{4}{3}x \pm 8 \).
Intersects x-axis (\( y=0 \)) at \( A = (6, 0) \).
Intersects y-axis (\( x=0 \)) at \( B = (0, 8) \).
Area of \( \triangle OAB = \frac{1}{2} \times 6 \times 8 = 24 \) sq. units.

Question. Let d be the perpendicular distance from the centre of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) to the tangent drawn at a point P on the ellipse. If \( F_1 \) & \( F_2 \) are the two foci of the ellipse, then show that \( (PF_1 – PF_2)^2 = 4a^2 \left[ 1 - \frac{b^2}{d^2} \right] \).
Answer: Let \( P = (x_1, y_1) \). Tangent: \( \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \).
Perpendicular distance from origin:
\( \frac{1}{d^2} = \frac{x_1^2}{a^4} + \frac{y_1^2}{b^4} \).
We know \( PF_1 = a + ex_1 \) and \( PF_2 = a - ex_1 \).
\( PF_1 - PF_2 = 2ex_1 \).
\( (PF_1 - PF_2)^2 = 4e^2x_1^2 \).
From the ellipse equation, \( \frac{y_1^2}{b^2} = 1 - \frac{x_1^2}{a^2} \).
Substitute into \( 1/d^2 \):
\( \frac{1}{d^2} = \frac{x_1^2}{a^4} + \frac{1}{b^2}(1 - \frac{x_1^2}{a^2}) = \frac{x_1^2 b^2 + a^2(a^2 - x_1^2)}{a^4 b^2} = \frac{a^4 - x_1^2(a^2 - b^2)}{a^4 b^2} = \frac{a^4 - x_1^2 a^2 e^2}{a^4 b^2} \).
\( \implies \) \( \frac{b^2}{d^2} = 1 - \frac{e^2 x_1^2}{a^2} \)
\( \implies \) \( \frac{e^2 x_1^2}{a^2} = 1 - \frac{b^2}{d^2} \)
\( \implies \) \( 4e^2 x_1^2 = 4a^2 (1 - \frac{b^2}{d^2}) \).
Thus, \( (PF_1 - PF_2)^2 = 4a^2 (1 - \frac{b^2}{d^2}) \).

Question. Common tangents are drawn to the parabola \( y^2 = 4x \) & the ellipse \( 3x^2 + 8y^2 = 48 \) touching the parabola at A & B and the ellipse at C & D. Find the area of the quadrilateral.
Answer: Ellipse: \( \frac{x^2}{16} + \frac{y^2}{6} = 1 \).
Tangent to parabola: \( y = mx + \frac{1}{m} \).
Tangent to ellipse: \( y = mx \pm \sqrt{16m^2 + 6} \).
Comparing: \( \frac{1}{m^2} = 16m^2 + 6 \)
\( \implies \) \( 16m^4 + 6m^2 - 1 = 0 \)
\( \implies \) \( (8m^2 - 1)(2m^2 + 1) = 0 \)
\( \implies \) \( m^2 = 1/8 \implies m = \pm \frac{1}{2\sqrt{2}} \).
Tangents: \( y = \frac{1}{2\sqrt{2}}x + 2\sqrt{2} \) and \( y = -\frac{1}{2\sqrt{2}}x - 2\sqrt{2} \).
Point of contact on parabola (\( a/m^2, 2a/m \)): \( A = (8, 4\sqrt{2}) \), \( B = (8, -4\sqrt{2}) \).
Point of contact on ellipse (\( -a^2m/c, b^2/c \)): \( C = (-4, 3/\sqrt{2}) \), \( D = (-4, -3/\sqrt{2}) \).
The quadrilateral ABCD is a trapezium. Area \( = \frac{1}{2} (8\sqrt{2} + 3\sqrt{2}) \times 12 = 66\sqrt{2} \).
*(Note: Re-checking coordinates for exact result as per solution sheet).*

Question. If the tangent at any point of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) makes an angle \( \alpha \) with the major axis and an angle \( \beta \) with the focal radius of the point of contact then show that the eccentricity ‘e’ of the ellipse is given by the absolute value of \( \frac{\cos \beta}{\cos \alpha} \).
Answer: Let point of contact be \( P(a \cos \theta, b \sin \theta) \).
Tangent slope \( \tan \alpha = -\frac{b \cos \theta}{a \sin \theta} \).
Slope of focal radius \( SP \) (\( S = ae, 0 \)): \( m_2 = \frac{b \sin \theta}{a \cos \theta - ae} \).
Using the formula for angle between lines \( \tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \).
After trigonometric substitution and simplification:
\( e = \left| \frac{\cos \beta}{\cos \alpha} \right| \).

Question. An ellipse has foci at \( F_1 \)(9, 20) and \( F_2 \)(49, 55) in the xy-plane and is tangent to the x-axis. Find the length of its major axis.
Answer: By reflection property of an ellipse, the reflection of one focus in any tangent line must lie on the circle with the other focus as centre and radius equal to the length of the major axis \( 2a \).
Tangent is x-axis (\( y=0 \)).
Reflection of \( F_1 \)(9, 20) in x-axis is \( F_1' \)(9, –20).
The distance \( F_2 F_1' = 2a \).
\( (2a)^2 = (49 - 9)^2 + (55 - (-20))^2 \)
\( (2a)^2 = 40^2 + 75^2 = 1600 + 5625 = 7225 \)
\( 2a = \sqrt{7225} = 85 \).
The length of the major axis is 85.

Advanced Subjective Questions

Question. PG is the normal to a standard ellipse at P, G being on the major axis. GP is produced outwards to Q so that PQ = GP. Show that the locus of Q is an ellipse whose eccentricity is \( \frac{a^2 - b^2}{a^2 + b^2} \).
Answer: Equation of normal to the ellipse \( ax \sec \theta - by \csc \theta = a^2 - b^2 \) .....(i)
so point \( P = (a \cos \theta, b \sin \theta) \)
point \( G = \left( \frac{a^2 - b^2}{a} \cos \theta, 0 \right) \)
Let point \( Q = (h, k) \)
so \( ah \sec \theta - bk \csc \theta = a^2 - b^2 \) .....(ii)
& Given PQ = GP
\( \implies (h - a \cos \theta)^2 + (k - b \sin \theta)^2 = \left( a \cos \theta - \frac{a^2 - b^2}{a} \cos \theta \right)^2 + (b \sin \theta)^2 \) .....(iii)
using (ii) & (iii), locus of Q is ellipse.

Question. P & Q are the corresponding points on a standard ellipse & its auxiliary circle. The tangent at P to the ellipse meets the major axis in T. Prove that QT touches the auxiliary circle.
Answer: Point \( P = (a \cos \theta, b \sin \theta) \)
& \( Q = (a \cos \theta, a \sin \theta) \)
equation of tangent to ellipse at 'P' \( bx \cos \theta + ay \sin \theta = ab \)
point \( T = (a \sec \theta, 0) \)
equation of line joining Q & T
\( y = \frac{a \sin \theta}{a \cos \theta - a \sec \theta} (x - a \sec \theta) \)
which is tangent to the circle \( x^2 + y^2 = a^2 \)

Question. A tangent to the ellipse \( x^2 + 4y^2 = 4 \) meets the ellipse \( x^2 + 2y^2 = 6 \) at P & Q. Prove that the tangents at P & Q of the ellipse \( x^2 + 2y^2 = 6 \) are at right angles.
Answer: Equation of tangent at \( (2 \cos \theta, \sin \theta) \)
\( x \cos \theta + 2y \sin \theta = 2 \) .......(1)
\( x^2 + 2y^2 = 6 \) .......(2)
Let a point is \( (x_1, y_1) \) on ellipse (2)
at which tangents intersects each other then equation of COC of \( (x_1, y_1) \) is
\( xx_1 + 2yy_1 - 6 = 0 \) .......(3)
equation (1) & (3) are same, so on comparing
\( x_1/3 = \cos \theta \) & \( y_1/3 = \sin \theta \implies x_1^2 + y_1^2 = 9 \)
locus of \( (x_1, y_1) \implies x^2 + y^2 = 9 \)
which is the director circle of ellipse (2), hence tangents cuts each other at 90°.

Question. Consider the parabola \( y^2 = 4x \) and the ellipse \( 2x^2 + y^2 = 6 \), intersecting at P and Q.
(a) Prove that the two curves are orthogonal.
(b) If tangent and normal at the point P on the ellipse intersect the x-axis at T and G respectively then find the area of the triangle PTG.
Answer: (a) \( c_1 \); \( y^2 = 4x \)
\( 2yy_1 = 4 \)
\( m_1 = y_1 = \frac{4}{2y} = \frac{2}{y} \)
\( c_2 \); \( 2x^2 + y^2 = 6 \)
\( 4x + 2yy_1 = 0 \)
\( m_2 = y_1 = -\frac{2x}{y} \)
\( m_1 m_2 = -\frac{4x}{y^2} = -\frac{4x}{4x} = -1 \)
curves are orthogonal.
(b) Tangent at P: \( y + x = 3 \implies T(3, 0) \)
Normal at P: \( x - y = -1 \implies G(-1, 0) \)
Area = \( \frac{1}{2} \times 4 \times 2 = 4 \)

Question. A normal inclined at 45° to the axis of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is drawn. It meets the x-axis & the y-axis in P & Q respectively. If C is the centre of the ellipse, show that the area of triangle CPQ is \( \frac{(a^2 - b^2)^2}{2(a^2 + b^2)} \) sq. units.
Answer: Equation of normal with slope 1
\( y = x - \frac{(a^2 - b^2)}{\sqrt{a^2 + b^2}} \) .....(i)
point \( P = \left( \frac{(a^2 - b^2)}{\sqrt{a^2 + b^2}}, 0 \right) \); \( Q = \left( 0, -\frac{(a^2 - b^2)}{\sqrt{a^2 + b^2}} \right) \)
\( C = (0, 0) \)
Area of \( \triangle CPQ = \frac{1}{2} \begin{vmatrix} \frac{a^2 - b^2}{\sqrt{a^2 + b^2}} & 0 & 1 \\ 0 & \frac{-(a^2 - b^2)}{\sqrt{a^2 + b^2}} & 1 \\ 0 & 0 & 1 \end{vmatrix} \)
\( = \frac{(a^2 - b^2)^2}{2(a^2 + b^2)} \) sq. units.

Question. Tangents are drawn to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) from the point \( \left( \frac{a^2}{\sqrt{a^2 - b^2}}, \sqrt{a^2 + b^2} \right) \). Prove that they intercept on the ordinate through the nearer focus a distance equal to the major axis.
Answer: Equation of tangent at \( \left( \frac{a^2}{\sqrt{a^2 - b^2}}, \sqrt{a^2 + b^2} \right) \)
\( \frac{x.a^2}{\sqrt{a^2 - b^2}.a^2} + \frac{y.\sqrt{a^2 + b^2}}{b^2} = 1 \)
\( \implies b^2 x + y \sqrt{a^4 - b^4} = b^2 \sqrt{a^2 - b^2} \) .....(i)
Intercept on x = ae

Question. A ray emanating from the point (– 4, 0) is incident on the ellipse \( 9x^2 + 25y^2 = 225 \) at the point P with abscissa 3. Find the equation of the reflected ray after first reflection.
Answer: For the ellipse \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \)
focii are (–4, 0) & (4, 0)
As we know that a ray emanating from one focus, after first reflection passes through the other focus.
so point on the ellipse are A(12/5, 3) & B(–12/5, 3)
so Reflecting ray will be equation of line joining (4, 0) & A and (4, 0) & B.

Question. If p is the length of the perpendicular from the focus ‘S’ of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) on any tangent at ‘P’, then show that \( \frac{b^2}{p^2} = \frac{2a}{\ell(SP)} - 1 \).
Answer: Tangent at \( P(x_1, y_1) \)
\( \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \)
Let S be (ae, 0) then \( p = \frac{\left| \frac{ae x_1}{a^2} - 1 \right|}{\sqrt{\frac{x_1^2}{a^4} + \frac{y_1^2}{b^4}}} \)
\( \implies \frac{1}{p^2} = \frac{x_1^2/a^4 + y_1^2 a^2/b^4}{(a - ex_1)^2} \)
\( \implies \frac{b^2}{p^2} = \frac{b^2 \left[ \frac{x_1^2}{a^4} + \frac{1}{a^2} \left( 1 - \frac{x_1^2}{a^2} \right) \right]}{(a - ex_1)^2} \)
\( \implies \frac{b^2}{p^2} = \frac{(a^2 - (1 - b^2/a^2)x_1^2)}{(a - ex_1)^2} \)
\( \implies \frac{b^2}{p^2} = \frac{a^2 - e^2 x_1^2}{(a - ex_1)^2} = \frac{a + ex_1}{a - ex_1} = \frac{2a}{\ell(SP)} - 1 \)
\( \because (\ell(SP) = \text{focal distance}) \)

Question. Variable pairs of chords at right angles and drawn through any point P (with eccentric angle \( \pi/4 \)) on the ellipse \( \frac{x^2}{4} + y^2 = 1 \), to meet the ellipse at two points say A and B. If the line joining A and B passes through a fixed point Q(a, b) such that \( a^2 + b^2 \) has the value equal to \( \frac{m}{n} \), where m, n are relatively prime positive integers, find (m + n).
Answer: Point \( P = (\sqrt{2}, 1/\sqrt{2}) \)
shifting the ellipse by letting the origin at \( (\sqrt{2}, 1/\sqrt{2}) \)
\( (x + \sqrt{2})^2 + 4(y + 1/\sqrt{2})^2 = 4 \)
\( \implies x^2 + 4y^2 + 2\sqrt{2}x + 8\sqrt{2}y = 0 \) ....(1)
Let the line AB \( \ell x + my = 1 \) ....(2)
Homozining (1) with (2) & as the angle between the chords is 90° so coeff. of \( x^2 \) + coeff. of \( y^2 \) = 0
\( \implies 2\sqrt{2} \ell + 4\sqrt{2} m = -5 \) ....(3)
using (2) & (3) \( \left( \frac{-5}{2\sqrt{2}}x - 1 \right) + m(y - 2x) = 0 \) ....(4)
which shows a family of line & passes through a fixed point which is point of intersection of two line A.
\( \implies x = -\frac{2\sqrt{2}}{5} \) & \( y = -\frac{4\sqrt{2}}{5} \)
again \( x = -\frac{2\sqrt{2}}{5} - \sqrt{2} = -\frac{3\sqrt{2}}{5} \) & \( y = \frac{3\sqrt{2}}{10} \)
\( a^2 + b^2 = \frac{9}{10} \)
\( \implies a + b = 19 \)