If the line x - 2y = 12 is a tangent to the ellipse x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1 at the point (3, - (9)/(2)) , then the length of the latus rectum of the ellipse is

Correct answer: C. 9 units ∵ (3, - (9)/(2)) lies on x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1 ⇒ 9 a ^ 2 + 81 4 b ^ 2 = 1 ……(1) Equation of the tangent at (3, - (9)/(2)) is 3x a ^ 2 + - (9)/(2)y b ^ 2 = 1 & given equation of the tangent is: x - 2y = 12 ⇒ (x)/(12) + (- y)/(6) = 1 On comparing these equations: a ^ 2 3 = 12 ⇒ a ^ 2 = 36 ⇒ a = 6 2 b ^ 2 9 = 6 ⇒ b ^ 2 = 27 ⇒ b = 3 3 Therefore, the length of latus rectum = 2 b ^ 2 a = (2 × 27)/(6) = 9

JEE Main 201910 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Ellipse

If the line x - 2y = 12 is a tangent to the ellipse x ^ 2 a ^ 2 + y...

If the line

x - 2y = 12

is a tangent to the ellipse

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

at the point \left(3, - \dfrac{9}{2}\right)$, then the length of the latus rectum of the ellipse is

Options

A $5$ units
B $12\sqrt{2}$ units
C $9$ units
D $8\sqrt{3}$ units

Correct answer

C. $9$ units

Step-by-step solution

∵

\left(3, - \dfrac{9}{2}\right)

lies on

\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1 ⇒ \dfrac{9}{{a}^{2}} + \dfrac{81}{4{b}^{2}} = 1

……\left(1\right)

<p>Equation of the tangent at \left(3, - \dfrac{9}{2}\right)

\dfrac{3x}{{a}^{2}} + \dfrac{- \dfrac{9}{2}y}{{b}^{2}} = 1

& given equation of the tangent is:

$x - 2y = 12 ⇒ \dfrac{x}{12} + \dfrac{- y}{6} = 1$

On comparing these equations:

$\dfrac{{a}^{2}}{3} = 12 ⇒ {a}^{2} = 36 ⇒ a = 6$

$\dfrac{2{b}^{2}}{9} = 6 ⇒ {b}^{2} = 27 ⇒ b = 3\sqrt{3}$

Therefore, the length of latus rectum

$= \dfrac{2{b}^{2}}{a} = \dfrac{2 \times 27}{6} = 9$

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