If 3x + 4y = 12 2 is a tangent o the ellipse x ^ 2 a ^ 2 + y ^ 2 9 = 1 for some a ∈ R, then the distance between the foci of the ellipse is

Correct answer: A. 2 7 3x + 4y = 12 2 ⇒ 4y = - 3x + 12 2 ⇒ y = - (3)/(4)x + 3 2 Condition of tangency c ^ 2 = a ^ 2 m ^ 2 + b ^ 2 18 = a ^ 2 ∙ (9)/(16) + 9 a ^ 2 ∙ (9)/(16) = 9 a ^ 2 = 16 a = 4 e = 1 - b ^ 2 a ^ 2 = 1 - (9)/(16) = 7 4 ∴ ae = 7 4 ∙ 4 = 7 ∴ Focus are (± 7 , 0) ∴ Distance between foci = 2 7

JEE Main 20207 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2020 · 7 January · Evening Shift · Mathematics · Ellipse

If 3x + 4y = 12 2 is a tangent o the ellipse x ^ 2 a ^ 2 + y ^ 2 9...

3x + 4y = 12\sqrt{2}

is a tangent o the ellipse

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

for some

a \in R,

then the distance between the foci of the ellipse is

Options

A $2\sqrt{7}$
B $4$
C $2\sqrt{5}$
D $2\sqrt{2}$

Correct answer

A. $2\sqrt{7}$

Step-by-step solution

3x + 4y = 12\sqrt{2}

<br />

⇒ 4y = - 3x + 12\sqrt{2}

<br />

⇒ y = - \dfrac{3}{4}x + 3\sqrt{2}

<br />Condition of tangency

{c}^{2} = {a}^{2}{m}^{2} + {b}^{2}

<br />

18 = {a}^{2} ∙ \dfrac{9}{16} + 9

<br />

{a}^{2} ∙ \dfrac{9}{16} = 9

<br />

{a}^{2} = 16

<br />

a = 4

<br />

e = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}} = \sqrt{1 - \dfrac{9}{16}} = \dfrac{\sqrt{7}}{4}

<br />

∴ ae = \dfrac{\sqrt{7}}{4} ∙ 4 = \sqrt{7}

<br />

∴

Focus are \left(\pm \sqrt{7}, 0\right)

∴

Distance between foci

= 2\sqrt{7}

Practice Ellipse on Quantrex Academy →

More from Ellipse