The equation of the tangent of the ellipse $4 x^2+9 y^2=36$ at the end of the latusrectum lying in the…
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
$\sqrt{5} x-3 y+1=0$
$x-3 y+\sqrt{5}=0$
$\sqrt{5} x-3 y+3=0$
$\sqrt{5} x-3 y+9=0$
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}
$