Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Equation of tangents to ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ , which are perpendicular to the line 3x + 4y= -7

Option 1)
$4x-3y=\pm 6\sqrt{5}$

Option 2)
$4x-3y=\pm \sqrt{12}$

Option 3)
$4x-3y=\pm \sqrt{2}$

Option 4)
$4x-3y=\pm {1}$

Answers (1)

best_answer

Standard equation -

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

- wherein

$a\rightarrow$ Semi major axis

$b\rightarrow$ Semi minor axis

Slope of tangent $=\frac{-1}{\left ( \frac{-3}{4} \right )} = \frac{4}{3}$

So, equation of tangent at $(3\cos \theta, 2\sin \theta )$

$\frac{\cos \theta x}{3}+\frac{\sin \theta y}{2} = 1$

where, $\frac{-\cos \theta ^{2}}{3\sin \theta } =\frac{4}{3}$

$\Rightarrow \tan \theta = \frac{-1}{2} \Rightarrow \sin \theta = \frac{1}{\sqrt{5}}$ &

$\cos \theta =\frac{-2}{\sqrt{5}}$

$\therefore$ equation $\frac{-2}{\sqrt{5}3}x+\frac{1}{\sqrt{5}2}y=1$

$\Rightarrow -4x+3y=6\sqrt{5}$

$\Rightarrow 4x-3y=6\sqrt{5}$

Option 1)

$4x-3y=\pm 6\sqrt{5}$

This solution is correct

Option 2)

$4x-3y=\pm \sqrt{12}$

This solution is incorrect

Option 3)

$4x-3y=\pm \sqrt{2}$

This solution is incorrect

Option 4)

$4x-3y=\pm {1}$

This solution is incorrect

Posted by

Aadil

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 registration window closes tomorrow; exam from May 20

anam-khan

BITSAT 2024 registration deadline extended till April 16; session 1 exam from May 19 to 24

Latest Question