JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Ellipse

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Let the length of the latus rectum of an ellipse with its major axis along x-axis and center at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?

(A) $(42, 22)$
(B) $(43, 22)$ ✓ Correct
(C) $(43, 23)$
(D) $(42, 23)$

Answer: (B)

Solution

For

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

with

a > b

, the latus rectum

\frac{2 b ^{2}}{a} = 8

gives

b^{2} = 4 a

, and the condition that the distance between foci equals the minor axis,

2 a e = 2 b

, gives

a e = b

. With

b^{2} = a^{2} (1 - e^{2})

and

a e = b

a^{2} e^{2} = a^{2} (1 - e^{2})

, so

e^{2} = \frac{1}{2}

; then

b^{2} = \frac{a ^{2}}{2}

, and with

b^{2} = 4 a

this gives

a = 8

b^{2} = 32

. The ellipse

\frac{x ^{2}}{64} + \frac{y ^{2}}{32} = 1

is satisfied by

(43, 22)

, since

\frac{48}{64} + \frac{8}{32} = \frac{3}{4} + \frac{1}{4} = 1

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Ellipse

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