ML Aggarwal Class 12 Maths Solutions Section A Chapter 04 Conics

Conics

Introduction

Curves called conics received their name because they form when a plane cuts through a right circular cone. A right circular cone is made when a line rotates around a circle and passes through a fixed point on the normal line through the circle's center. The fixed point V is the vertex of the cone and splits it into two sections known as nappes. Each line that moves through V is called a generator. The circle around which the generators rotate is called the base, and the line through the base's center and V is called the axis.

Around 200 B.C., Apollonius discovered that a conic section forms when a plane cuts a right circular cone with two nappes. This method produces three types of curves: parabola, hyperbola, and ellipse.

When a plane intersects a right circular cone, the resulting curve is:

(i) A parabola - when the cutting plane is parallel to exactly one generator of the cone. The plane cuts only one nappe.

(ii) A hyperbola - when the cutting plane is parallel to two generators of the cone. The plane cuts both nappes.

(iii) An ellipse - when the cutting plane is not parallel to any generator of the cone. The plane cuts only one nappe and crosses every generator.

4.1 Conic

A conic is now defined analytically. Let l be a fixed line and F be a fixed point not on l. Let e > 0 be a fixed real number. If |MP| represents the perpendicular distance from a point P to the line l, then the set of all points P where |FP| = e|MP| forms a conic.

The point F is known as a focus of the conic. The line l is known as the directrix linked with F. The constant e is called the eccentricity.

Depending on the value of eccentricity, a conic is classified as:

(i) A parabola - when e = 1
(ii) An ellipse - when e < 1
(iii) A hyperbola - when e > 1

4.2 Symmetry

Reflection of a point in a line

If a point P is on a line l, then its reflection in l is the point P itself. When P is not on l, we find the foot of the perpendicular from P to l (call it M) and extend it to a point P' where |MP| = |MP'|. This point P' is the reflection of P in l.

When P' is the reflection of P in l, then P is also the reflection of P' in l. Additionally, if P does not lie on l, then P' is the reflection of P in l if and only if l is the perpendicular bisector of segment PP'.

For a point (α, β), its reflection in:

(i) The origin is (-α, -β)
(ii) The x-axis is (α, -β)
(iii) The y-axis is (-α, β)

Reflection of a point in a point

The reflection of a point P in a fixed point M is the point P' where M is the midpoint of segment PP'.

If P = M, then P' = M = P.

If P' is the reflection of P in point M, then P is the reflection of P' in M.

Symmetry of a curve about a line

A curve C has line symmetry about a line l when for every point P on C, the reflection of P in l also lies on C. The line l is then called a line of symmetry or an axis of C.

For a curve with equation F(x, y) = 0, the curve is symmetric about:

(i) The x-axis if F(x, y) = F(x, -y)
(ii) The y-axis if F(x, y) = F(-x, y)

Symmetry of a curve about a point

A curve C shows point symmetry about a point M when for every point P on C, the reflection of P in M also lies on C. The point M is called a centre of symmetry or a centre of C.

For a curve with equation F(x, y) = 0, the curve is symmetric about the origin if F(x, y) = F(-x, -y).

Illustrative Example

Example. Which of the following curves are symmetrical about the x-axis or y-axis or origin?
(i) 7y² - 5x + 2 = 0
(ii) 3x² - 4y² + 7 = 0

Solution. (i) The equation of the given curve is F(x, y) = 7y² - 5x + 2 = 0.

Here, F(x, -y) = 7(-y)² - 5x + 2 = 7y² - 5x + 2 = F(x, y);
F(-x, y) = 7y² - 5(-x) + 2 = 7y² + 5x + 2 ≠ F(x, y);
F(-x, -y) = 7(-y)² - 5(-x) + 2 = 7y² + 5x + 2 ≠ F(x, y).

Therefore, the given curve is symmetric about the x-axis but not about the y-axis or origin.

(ii) The equation of the given curve is F(x, y) = 3x² - 4y² + 7 = 0.

Here, F(x, -y) = 3x² - 4(-y)² + 7 = 3x² - 4y² + 7 = F(x, y);
F(-x, y) = 3(-x)² - 4y² + 7 = 3x² - 4y² + 7 = F(x, y);
F(-x, -y) = 3(-x)² - 4(-y)² + 7 = 3x² - 4y² + 7 = F(x, y).

Therefore, the given curve is symmetric about the x-axis, y-axis, and origin.

Exercise 4.1

Question 1. Which of the following curves are symmetrical about the x-axis or y-axis or origin?
(i) 5x² + 7y - 6 = 0
(ii) y² = x + 9
(iii) x² = 3y² + 7
(iv) x² + y² = 25
(v) xy = 1
(vi) x² + y² - 4xy + 3 = 0
Answer: (i) Symmetric about y-axis only; (ii) Symmetric about x-axis only; (iii) Symmetric about x-axis, y-axis and origin; (iv) Symmetric about x-axis, y-axis and origin; (v) Symmetric about origin only; (vi) Symmetric about origin only.
In simple words: Check if replacing x with -x or y with -y gives back the same equation - that tells you which axis or point it's symmetric about.

Exam Tip: Always test all three symmetries (x-axis, y-axis, origin) for each curve before drawing your conclusion.

Question 2. If a curve is symmetrical about each of the co-ordinate axes then prove that it is also symmetrical about the origin. What can you say about the converse?
Answer: Let P(x, y) be a point on curve C. If C is symmetric about the x-axis, then (x, -y) also lies on C. If C is also symmetric about the y-axis, then (-x, -y) must lie on C. This means C is symmetric about the origin. For the converse, if a curve is symmetric about the origin, it need not be symmetric about both coordinate axes - a counterexample is xy - 1 = 0, which is symmetric about the origin but not about either axis.
In simple words: If a shape looks the same when you flip it across both the x and y axes, then it also looks the same when you spin it around the center. But spinning it around the center doesn't mean it flips the same way across the axes.

Exam Tip: For this proof-based question, show the logical chain clearly: first point on curve and symmetric about x-axis, then apply y-axis symmetry to get the origin symmetry.

4.3 Parabola

4.3.1 To find the equation of a parabola in the standard form y² = 4ax, a > 0

Let F be the focus and l be the directrix. Let Z be the foot of perpendicular from F to l. Take the line ZF as the x-axis with positive direction pointing from Z toward F. Let A be the midpoint of ZF, and place A at the origin. Then the line through A perpendicular to ZF becomes the y-axis.

Let |ZF| = 2a (where a > 0, because F does not lie on l). Then |ZA| = |AF| = a.

Since F is to the right of A and Z is to the left of A, the coordinates are F(a, 0) and Z(-a, 0).

The directrix equation is x = -a, or x + a = 0.

For any point P(x, y) in the plane, P lies on the parabola if and only if |FP| = |MP|, where |MP| is the perpendicular distance from P to the directrix. This gives:

\( \sqrt{(x - a)^2 + y^2} = \frac{|x + a|}{1} \)

Squaring both sides:

\( (x - a)^2 + y^2 = (x + a)^2 \)

\( x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2 \)

\( y^2 = 4ax \)

Therefore, the standard form of a parabola is y² = 4ax (a > 0), with focus F(a, 0) and directrix x + a = 0. This is also called the first standard form or right-hand parabola.

4.3.2 To determine the shape of the parabola y² = 4ax, a > 0

The equation is F(x, y) = y² - 4ax = 0.

The following facts describe the parabola:

1. F(x, -y) = (-y)² - 4ax = y² - 4ax = F(x, y), so the parabola is symmetric about the x-axis.

2. When x < 0, the equation y² = 4ax has no real solutions for y, so no points exist on the curve with negative x-coordinates (to the left of the y-axis). At x = 0, y² = 0 gives y = 0, so (0, 0) is the only point where the y-axis meets the curve. The entire curve except the origin lies to the right of the y-axis.

3. For a point P(x, y) on the parabola in the first quadrant, y = 2√(ax). As x increases, y also increases and the curve extends without bound. Sketch a portion in the first quadrant, then use symmetry to complete the full curve.

4.3.3 Some facts about the parabola y² = 4ax, a > 0

1. It is symmetric about the x-axis. This line is called the axis of the parabola.

2. Focus is F(a, 0) and directrix is x + a = 0.

3. The vertex A(0, 0) is where the axis meets the parabola. It is the midpoint of ZF, where F(a, 0) is the focus and Z is the foot of perpendicular from the focus to the directrix.

4. The curve lies entirely to the right of the y-axis, except at the origin.

5. The y-axis meets the curve only at the origin.

6. The curve is unbounded.

7. Every line parallel to the axis, i.e. y = c (c any real number), meets the parabola at exactly one point: \( \left( \frac{c^2}{4a}, c \right) \)

8. A line segment joining two points on the parabola is called a chord of the parabola.

9. A focal chord is any chord passing through the focus F. The focal distance of a point is its distance from the focus.

10. The latus rectum is a chord through the focus F and perpendicular to the axis of the parabola. Its length is called the length of the latus rectum.

11. Length of latus rectum: Let L'L be the latus rectum, passing through F(a, 0) perpendicular to the x-axis. If |FL| = l (> 0), then L and L' are at (a, l) and (a, -l) respectively. Since L(a, l) lies on y² = 4ax, we get l² = 4a·a, so l = 2a. Thus L(a, 2a) and L'(a, -2a). The length of latus rectum = |L'L| = 2l = 4a. The endpoints are L(a, 2a) and L'(a, -2a), and the equation of the latus rectum is x - a = 0.

4.3.4 To find the equation of a parabola in other standard forms

Find the equation of a parabola with:

(i) Focus F(-a, 0), a > 0 and directrix x - a = 0.

Using the condition |FP| = |MP|:

\( \sqrt{(x + a)^2 + y^2} = |x - a| \)

Squaring and simplifying:

\( (x + a)^2 + y^2 = (x - a)^2 \)

\( x^2 + 2ax + a^2 + y^2 = x^2 - 2ax + a^2 \)

\( y^2 = -4ax, \text{ } a > 0 \)

This is called the 2nd standard form or left-hand parabola.

(ii) Focus F(0, a), a > 0 and directrix y + a = 0.

By similar calculation: \( x^2 = 4ay, \text{ } a > 0 \)

This is called the 3rd standard form or upward parabola.

(iii) Focus F(0, -a), a > 0 and directrix y - a = 0.

By similar calculation: \( x^2 = -4ay, \text{ } a > 0 \)

This is called the 4th standard form or downward parabola.

Main facts about the parabola

Illustrative Examples

Example 1. Find the equation of the parabola with focus at (-2, 0) and whose directrix is the line x + 2y - 3 = 0.
Answer: The focus is F(-2, 0) and the directrix is x + 2y - 3 = 0. Let P(x, y) be any point on the parabola. Using the parabola condition |FP| = |MP|:

\( \sqrt{(x + 2)^2 + y^2} = \frac{|x + 2y - 3|}{\sqrt{1^2 + 2^2}} \)

Squaring both sides:

\( 5[(x + 2)^2 + y^2] = (x + 2y - 3)^2 \)

\( 5(x^2 + 4x + 4 + y^2) = x^2 + 4y^2 + 9 + 4xy - 6x - 12y \)

\( 4x^2 - 4xy + y^2 + 26x + 12y + 11 = 0 \)

This is the required equation of the parabola.
In simple words: Every point on this parabola is equally far from the focus (-2, 0) and the line x + 2y - 3 = 0.

Exam Tip: Always apply the distance formula from a point to a line carefully: \( \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \), and square both sides after setting distances equal.

Example 2. Find the focus, directrix and eccentricity of the conic represented by the equation 3y² = 8x.
Answer: Rewrite as \( y^2 = \frac{8}{3}x \), which matches the form \( y^2 = 4ax \). Comparing: \( 4a = \frac{8}{3} \), so \( a = \frac{2}{3} \). The eccentricity of a parabola is always e = 1. The focus is at \( (a, 0) = \left( \frac{2}{3}, 0 \right) \). The directrix is \( x + a = 0 \), or \( x + \frac{2}{3} = 0 \), which gives \( 3x + 2 = 0 \).
In simple words: Rewrite the equation in standard form, read off the value of a, and then use it to find the focus and directrix directly.

Exam Tip: Always remember that for any parabola, e = 1. Once you find the value of a, you can immediately state the focus and directrix.

Example 3. Find the focus, directrix and eccentricity of the conic represented by the equation 5x² = -12y.
Answer: Rewrite as \( x^2 = -\frac{12}{5}y \), which matches the form \( x^2 = -4ay \). Comparing: \( 4a = \frac{12}{5} \), so \( a = \frac{3}{5} \). The eccentricity is e = 1 (all parabolas). The focus is at \( (0, -a) = \left( 0, -\frac{3}{5} \right) \). The directrix is \( y - a = 0 \), or \( y - \frac{3}{5} = 0 \), which gives \( 5y - 3 = 0 \).
In simple words: This parabola opens downward. Rewrite to standard form, find a, and use it to get the focus on the y-axis and the directrix as a horizontal line.

Exam Tip: Notice that x² = -4ay (downward parabola) has the focus below the origin on the y-axis and the directrix above.

Example 4. Find the equation of the parabola with vertex at origin and directrix the line y + 3 = 0. Also find its focus.
Answer: The vertex is at the origin (0, 0). The directrix y + 3 = 0 means y = -3, a horizontal line 3 units below the origin. Since the vertex is equidistant from focus and directrix, and lies between them, the focus is 3 units above the origin at F(0, 3). The parabola opens upward and has the form \( x^2 = 4ay \) with a = 3. Therefore, the equation is \( x^2 = 12y \) and the focus is F(0, 3).
In simple words: When the directrix is horizontal and below the vertex, the parabola opens upward. The focus is the same distance above the vertex as the directrix is below it.

Exam Tip: Always identify whether the directrix is horizontal or at an angle. For horizontal or vertical directrices, you can use the standard forms directly.

Example 5. Find the equation of the parabola with vertex at origin, axis along x-axis and passing through the point (-1, 3).
Answer: The vertex is at the origin, the axis is along the x-axis, and the parabola passes through (-1, 3) in the second quadrant. This means the parabola opens to the left, so it has the form \( y^2 = -4ax \). Substitute the point (-1, 3): \( 9 = -4a(-1) \), so \( 4a = 9 \). The equation is \( y^2 = -9x \).
In simple words: Since the point is in the second quadrant (negative x, positive y), the parabola must open leftward. Substitute to find the value of a.

Exam Tip: Before writing the general form, always determine the correct standard form by checking which quadrant the given point lies in.

Example 6. Find the equation of the parabola with vertex at origin, symmetric with respect to y-axis and passing through (2, -3).
Answer: The vertex is at the origin and the parabola is symmetric about the y-axis, so the axis lies along the y-axis. The point (2, -3) is in the fourth quadrant (positive x, negative y), so the parabola opens downward. The form is \( x^2 = -4ay \). Substitute (2, -3): \( 4 = -4a(-3) \), so \( 4 = 12a \), giving \( a = \frac{1}{3} \). The equation is \( x^2 = -\frac{4}{3}y \), or \( 3x^2 = -4y \).
In simple words: Symmetry about the y-axis means the axis is the y-axis itself, so the parabola opens either up or down. The fourth-quadrant point tells us it opens downward.

Exam Tip: "Symmetric with respect to the y-axis" means the axis of the parabola IS the y-axis, not just that it's symmetric in general.

Example 7. Find the equation of the parabola whose vertex is at the point (4, 1) and focus at the point (6, -3).
Answer: The vertex is A(4, 1) and focus is F(6, -3). The line AF is the axis of the parabola. Since A is the midpoint between the focus F and the foot of perpendicular Z on the directrix, we find Z: if Z = (α, β), then \( \frac{\alpha + 6}{2} = 4 \) and \( \frac{\beta - 3}{2} = 1 \), giving α = 2, β = 5. So Z = (2, 5). The slope of the axis is \( \frac{-3 - 1}{6 - 4} = \frac{-4}{2} = -2 \). The directrix is perpendicular to this axis and passes through Z(2, 5), so its slope is \( \frac{1}{2} \). The directrix equation is \( y - 5 = \frac{1}{2}(x - 2) \), or \( x - 2y + 8 = 0 \). Using |FP| = |MP|:

\( \sqrt{(x - 6)^2 + (y + 3)^2} = \frac{|x - 2y + 8|}{\sqrt{1^2 + (-2)^2}} \)

Squaring:

\( 5[(x - 6)^2 + (y + 3)^2] = (x - 2y + 8)^2 \)

\( 4x^2 + 4xy + y^2 - 76x + 62y + 161 = 0 \)

This is the required equation.
In simple words: Use the fact that the vertex is the midpoint of the focus and the directrix's foot. Find the directrix using perpendicularity, then apply the parabola definition.

Exam Tip: When the vertex is not at the origin, locate the directrix first using the perpendicularity condition and the midpoint property. This simplifies the final calculation.

Example 8. The two lines ty = x + t² and y + tx = 2t + t³ intersect at the point P. Show that P lies on the curve whose equation is y² = 4x. (I.S.C. 2001)
Answer: The two given lines are:
\( ty - x = t^2 \) ...(i)
\( y + tx = 2t + t^3 \) ...(ii)

To find the locus, eliminate the parameter t. Multiply (i) by t and add to (ii):
\( t(ty - x) + (y + tx) = t \cdot t^2 + 2t + t^3 \)
\( t^2 y - tx + y + tx = t^3 + 2t + t^3 \)
\( (t^2 + 1)y = 2t^3 + 2t \)
\( y = \frac{2t(t^2 + 1)}{t^2 + 1} = 2t \)

So \( t = \frac{y}{2} \) ...(iii)

Substitute into (i):
\( \frac{y}{2} \cdot y - x = \left( \frac{y}{2} \right)^2 \)
\( \frac{y^2}{2} - x = \frac{y^2}{4} \)
\( \frac{y^2}{4} = x \)
\( y^2 = 4x \)

Hence, point P lies on the curve \( y^2 = 4x \).
In simple words: Eliminate t by combining the two equations algebraically to get a relationship between x and y that doesn't contain t.

Exam Tip: When eliminating a parameter from two line equations, try combining them by multiplication and addition to cancel out the parameter terms efficiently.

Exercise 4.2

Question 1. Write the equation of the parabola with the line x + y = 0 as directrix and the point (1, 0) as focus.
Answer: Let P(x, y) be any point on the parabola. The perpendicular distance from P to the line x + y = 0 is \( \frac{|x + y|}{\sqrt{2}} \). The distance from P to the focus (1, 0) is \( \sqrt{(x - 1)^2 + y^2} \). By the parabola definition:

\( \sqrt{(x - 1)^2 + y^2} = \frac{|x + y|}{\sqrt{2}} \)

Squaring both sides:

\( 2[(x - 1)^2 + y^2] = (x + y)^2 \)
\( 2[x^2 - 2x + 1 + y^2] = x^2 + 2xy + y^2 \)
\( 2x^2 - 4x + 2 + 2y^2 = x^2 + 2xy + y^2 \)
\( x^2 - 2xy + y^2 - 4x + 2 = 0 \)

This is the required equation of the parabola.
In simple words: Apply the distance formula for each point to both the focus and the directrix line, then set them equal per the parabola definition.

Exam Tip: Remember the perpendicular distance from point (x, y) to line ax + by + c = 0 is \( \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} \).

Question 2. Find the equation of the parabola having focus at (3, -4) and directrix as x + y = 2. (I.S.C. 2007)
Answer: Let P(x, y) be any point on the parabola. The distance from P to focus (3, -4) is \( \sqrt{(x - 3)^2 + (y + 4)^2} \). The perpendicular distance to directrix x + y - 2 = 0 is \( \frac{|x + y - 2|}{\sqrt{2}} \). By definition:

\( \sqrt{(x - 3)^2 + (y + 4)^2} = \frac{|x + y - 2|}{\sqrt{2}} \)

Squaring:

\( 2[(x - 3)^2 + (y + 4)^2] = (x + y - 2)^2 \)
\( 2[x^2 - 6x + 9 + y^2 + 8y + 16] = x^2 + y^2 + 4 + 2xy - 4x - 4y \)
\( 2x^2 - 12x + 18 + 2y^2 + 16y + 32 = x^2 + y^2 + 4 + 2xy - 4x - 4y \)
\( x^2 - 2xy + y^2 - 8x + 20y + 46 = 0 \)

This is the required equation.
In simple words: Find the distance from any point to the focus and to the directrix line, then use the defining property that these distances are equal for all parabola points.

Exam Tip: Always simplify the equation by squaring, but check your algebra carefully since expansions can be lengthy.

Question 3. Find the equation to the parabola whose focus is (-2, 1) and directrix is 6x - 3y = 8.
Answer: Let P(x, y) be a point on the parabola. By definition, |FP| = |MP| where |MP| is the distance to the directrix. The distance from P to F(-2, 1) is \( \sqrt{(x + 2)^2 + (y - 1)^2} \). The distance from P to line 6x - 3y - 8 = 0 is \( \frac{|6x - 3y - 8|}{\sqrt{36 + 9}} = \frac{|6x - 3y - 8|}{\sqrt{45}} = \frac{|6x - 3y - 8|}{3\sqrt{5}} \). Setting distances equal and squaring:

\( (x + 2)^2 + (y - 1)^2 = \frac{(6x - 3y - 8)^2}{45} \)
\( 45[(x + 2)^2 + (y - 1)^2] = (6x - 3y - 8)^2 \)

Expanding and simplifying yields:

\( 9x^2 + 36xy + 36y^2 + 276x - 138y - 161 = 0 \)

This is the required equation.
In simple words: Calculate the distance to an oblique directrix using the formula for distance from a point to a line, then apply the parabola condition.

Exam Tip: When the directrix is not parallel to a coordinate axis, be extra careful with the distance formula - include all coefficients in the denominator.

Question 4. The equation of the directrix of the parabola is 3x + 2y + 1 = 0. The focus is (2, 1). Find the equation of the parabola. (I.S.C. 2002)
Answer: Let P(x, y) be on the parabola. Distance to focus (2, 1) is \( \sqrt{(x - 2)^2 + (y - 1)^2} \). Distance to directrix 3x + 2y + 1 = 0 is \( \frac{|3x + 2y + 1|}{\sqrt{9 + 4}} = \frac{|3x + 2y + 1|}{\sqrt{13}} \). By definition:

\( \sqrt{(x - 2)^2 + (y - 1)^2} = \frac{|3x + 2y + 1|}{\sqrt{13}} \)

Squaring:

\( 13[(x - 2)^2 + (y - 1)^2] = (3x + 2y + 1)^2 \)
\( 13[x^2 - 4x + 4 + y^2 - 2y + 1] = 9x^2 + 4y^2 + 1 + 12xy + 6x + 4y \)
\( 4x^2 - 12xy + 9y^2 - 58x - 30y + 64 = 0 \)

This is the required equation.
In simple words: Use the standard parabola property: any point is equally distant from the focus and the directrix line.

Exam Tip: Always double-check the denominator in the distance-to-line formula by computing \( \sqrt{a^2 + b^2} \) for the line ax + by + c = 0.

4.4 Ellipse

4.4.1 To find the equation of an ellipse in the standard form

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \text{ where } b^2 = a^2(1 - e^2) \]

Let F be a focus, l be a directrix, and Z be the foot of perpendicular from F to l. Take FZ as the x-axis with positive direction from F to Z. Divide segment FZ internally and externally in ratio e : 1 at points A and A' respectively, where:

\( FA = e \cdot AZ \) and \( A'F = e \cdot A'Z \)

Since 0 < e < 1, point A' is closer to F and lies to the left of F. Let O be the midpoint of A'A. Take O as origin, and the line through O perpendicular to FZ becomes the y-axis.

Let A'A = 2a (a > 0), so A'O = OA = a. Points A and A' are (a, 0) and (-a, 0) respectively.

From the relationships above, we can show that the focus F is at (ae, 0) and the directrix is \( x = \frac{a}{e} \), or \( x - \frac{a}{e} = 0 \).

For any point P(x, y) on the ellipse, the defining condition is |FP| = e|MP|, where 0 < e < 1 and |MP| is the perpendicular distance to the directrix. This gives:

\( \sqrt{(x - ae)^2 + y^2} = e \left| x - \frac{a}{e} \right| \)

Squaring and simplifying:

\( (x - ae)^2 + y^2 = e^2 \left( x - \frac{a}{e} \right)^2 \)

\( x^2 - 2aex + a^2e^2 + y^2 = e^2x^2 - 2aex + a^2 \)

\( (1 - e^2)x^2 + y^2 = a^2(1 - e^2) \)

\( \frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1 \)

\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \text{ where } b^2 = a^2(1 - e^2) \)