Question

The tangent to the ellipse $16x^2+9y^2=144$, making equal intercepts on both the axes, is

• A. $y=x+3$
• B. $y=x-2$
• C. $x+y=5$
• D. $y=-x+4$

Answer 1

The correct option is C $x+y=5$

Given ellipse is $16x^2+9y^2=144$
$\Rightarrow \frac{x^2}{9}+\frac{y^2}{16}=1$
So tangent to the ellipse at any point ${}^{\prime }{\theta }^{\prime }$ is given by,
$\frac{x\cos \theta }{3}+\frac{y\sin \theta }{4}=1$
$\therefore x-$intercept $=\frac{3}{\cos \theta }$ and $y-$intercept $=\frac{4}{\sin \theta }$
Now given both intercepts are equal $\Rightarrow \frac{3}{\cos \theta }=\frac{4}{\sin \theta }\Rightarrow \tan \theta =\frac{4}{3}\therefore \sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}$
therefore required equation of the tangent is, $x+y=5$