Q97 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics

The equation of the tangent of the ellipse $4 x^2+9 y^2=36$ at the end of the latusrectum lying in the second quadrant, is

A
$\sqrt{5} x-3 y+1=0$
B
$x-3 y+\sqrt{5}=0$
C
$\sqrt{5} x-3 y+3=0$
Correct answer
D
$\sqrt{5} x-3 y+9=0$

Solution

Equation of given ellipse is $$ 4 x^2+9 y^2=36 \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1 $$ Now, coordinate of end of the latus rectum lying in the second quadrant is $P\left(-\sqrt{5}, \frac{4}{3}\right)$. So, the equation of tangent at point $P$ is $$ \begin{array}{rlrl} -4 \sqrt{5} x+12 y & =36 \\ \Rightarrow \quad & \sqrt{5} x-3 y+9 & =0 \end{array} $$

Hence, the correct answer is option: 'D'.

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