Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is (1)/(2) . If P ( 1, β ) , β > 0 is a point on this ellipse, then the equation of the normal to it at P is

Correct answer: D. 4x – 2y = 1 (a)/(e) = 4 ⇒ a = 4e ⇒ a = 2 b ^ 2 = a ^ 2 (1 - e ^ 2 ) = 3 ( 1, β ) lies on x ^ 2 4 + y ^ 2 3 = 1 ⇒ (1)/(4) + β ^ 2 3 = 1 ⇒ β ^ 2 = (9)/(4) ⇒ β = (3)/(2) ( ∵ β > 0 ) Normal at ( 1, β ) ⇒ a ^ 2 x 1 - b ^ 2 y β = a ^ 2 - b ^ 2 ⇒ 4x - (3y)/(β) = 1 so equation of normal is 4x - 2y = 1

JEE Main 20204 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2020 · 4 September · Evening Shift · Mathematics · Ellipse

Let x = 4 be a directrix to an ellipse whose centre is at the...

Let

x = 4

be a directrix to an ellipse whose centre is at the origin and its eccentricity is

\dfrac{1}{2} .

P ( 1, \beta ) , \beta > 0

is a point on this ellipse, then the equation of the normal to it at

P

Options

A $4x – 3y = 2$
B $8x – 2y = 5$
C $7x – 4y = 1$
D $4x – 2y = 1$

Correct answer

D. $4x – 2y = 1$

Step-by-step solution

$\dfrac{a}{e} = 4 ⇒ a = 4e ⇒ a = 2$

${b}^{2} = {a}^{2}\left(1 - {e}^{2}\right) = 3$

$( 1, \beta ) \text{lies on}\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1 ⇒ \dfrac{1}{4} + \dfrac{{\beta}^{2}}{3} = 1 ⇒ {\beta}^{2} = \dfrac{9}{4} ⇒ \beta = \dfrac{3}{2} ( ∵ \beta > 0 )$

Normal at $( 1, \beta ) ⇒ \dfrac{{a}^{2}x}{1} - \dfrac{{b}^{2}y}{\beta} = {a}^{2} - {b}^{2} ⇒ 4x - \dfrac{3y}{\beta} = 1$

so equation of normal is $4x - 2y = 1$

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