JEE Challenger
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MathMedAdv 2008 • Paper 1

Equations of Parabolas Having Same Latus Rectum as Ellipse

Let P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2), y1<0y_1 < 0, y2<0y_2 < 0, be the end points of the latus rectum of the ellipse x2+4y2=4x^2 + 4y^2 = 4. The equations of parabolas with latus rectum PQPQ are

Options

A

x2+23 y=3+3x^2 + 2\sqrt{3}\ y = 3 + \sqrt{3}

B

x2−23 y=3+3x^2 - 2\sqrt{3}\ y = 3 + \sqrt{3}

Correct
C

x2+23 y=3−3x^2 + 2\sqrt{3}\ y = 3 - \sqrt{3}

Correct
D

x2−23 y=3−3x^2 - 2\sqrt{3}\ y = 3 - \sqrt{3}

Topics & Concepts

Conic SectionsParabolaEllipse

Step-by-Step Solution

To find the equations of the parabolas, we first determine the coordinates of the points PP and QQ.

The given ellipse is:
x2+4y2=4  ⟹  x24+y21=1x^2 + 4y^2 = 4 \implies \frac{x^2}{4} + \frac{y^2}{1} = 1
Here, a2=4  ⟹  a=2a^2 = 4 \implies a = 2 and b2=1  ⟹  b=1b^2 = 1 \implies b = 1.

The eccentricity ee of the ellipse is given by:
e=1−b2a2=1−14=32e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}

The foci of the ellipse lie on the xx-axis at (±ae,0)=(±3,0)(\pm ae, 0) = (\pm \sqrt{3}, 0).

The latera recta are the double ordinates passing through the foci, parallel to the yy-axis, given by x=±3x = \pm \sqrt{3}. Substituting x=±3x = \pm \sqrt{3} into the equation of the ellipse:
(±3)2+4y2=4  ⟹  3+4y2=4  ⟹  y2=14  ⟹  y=±12(\pm \sqrt{3})^2 + 4y^2 = 4 \implies 3 + 4y^2 = 4 \implies y^2 = \frac{1}{4} \implies y = \pm \frac{1}{2}

The problem specifies the endpoints with y1<0y_1 < 0 and y2<0y_2 < 0. Thus, the points are the lower endpoints of the latera recta:
P=(−3,−12)andQ=(3,−12)P = \left(-\sqrt{3}, -\frac{1}{2}\right) \quad \text{and} \quad Q = \left(\sqrt{3}, -\frac{1}{2}\right)

The line segment PQPQ is horizontal, lying along the line y=−12y = -\frac{1}{2}.
The length of this segment is:
Length of latus rectum (4a)=3−(−3)=23  ⟹  a=32\text{Length of latus rectum } (4a) = \sqrt{3} - (-\sqrt{3}) = 2\sqrt{3} \implies a = \frac{\sqrt{3}}{2}

The midpoint of the segment PQPQ is the focus SS of the parabola:
S=(−3+32,−12)=(0,−12)S = \left(\frac{-\sqrt{3} + \sqrt{3}}{2}, -\frac{1}{2}\right) = \left(0, -\frac{1}{2}\right)

Since the latus rectum PQPQ is horizontal, the axis of the parabola is vertical (the yy-axis, x=0x = 0).

There are two such parabolas:

  1. Parabola opening upwards:
    The vertex V1V_1 lies at distance aa below the focus SS:
    V1=(0,−12−a)=(0,−12−32)V_1 = \left(0, -\frac{1}{2} - a\right) = \left(0, -\frac{1}{2} - \frac{\sqrt{3}}{2}\right)
    The equation of the parabola is:
    x2=4a(y−yV1)x^2 = 4a\left(y - y_{V_1}\right)
    x2=23(y−(−12−32))=23y+3+3x^2 = 2\sqrt{3}\left(y - \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}\right)\right) = 2\sqrt{3}y + \sqrt{3} + 3
    Rearranging:
    x2−23y=3+3x^2 - 2\sqrt{3}y = 3 + \sqrt{3}

  2. Parabola opening downwards:
    The vertex V2V_2 lies at distance aa above the focus SS:
    V2=(0,−12+a)=(0,−12+32)V_2 = \left(0, -\frac{1}{2} + a\right) = \left(0, -\frac{1}{2} + \frac{\sqrt{3}}{2}\right)
    The equation of the parabola is:
    x2=−4a(y−yV2)x^2 = -4a\left(y - y_{V_2}\right)
    x2=−23(y−(−12+32))=−23y−3+3x^2 = -2\sqrt{3}\left(y - \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\right)\right) = -2\sqrt{3}y - \sqrt{3} + 3
    Rearranging:
    x2+23y=3−3x^2 + 2\sqrt{3}y = 3 - \sqrt{3}

Therefore, the correct options are B and C.

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