Question

Find the equation of tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at the point whose eccentric angle is $\frac{4\pi }{3}$

• A. $4x+5\sqrt{3}y-40=0$
• B. None of the above
• C. $4x+5\sqrt{3}y+40=0$
• D. $8x+5\sqrt{3}y+80=0$

Answer 1

The correct option is C $4x+5\sqrt{3}y+40=0$

Given equation of ellipse is
$\frac{x^2}{25}+\frac{y^2}{16}=1$
On comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$a=5\text{and}b=4$
Equation of tangent is
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Given eccentric angle is $\frac{4\pi }{3}$
So, $\cos \left(\frac{4\pi }{3}\right)=-\frac{1}{2},\sin \left(\frac{4\pi }{3}\right)=-\frac{\sqrt{3}}{2}$

$\therefore \frac{x(-\frac{1}{2})}{5}+\frac{x(-\frac{\sqrt{3}}{2})}{4}=1$
$\Rightarrow -\frac{x}{10}-\frac{\sqrt{3}y}{8}=1$
$\Rightarrow 4x+5\sqrt{3}y=-40$
$\Rightarrow 4x+5\sqrt{3}y+40=0$
This is the required equation of tangent at $\frac{4\pi }{3}$