Question

Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at the points $A$ and $B$.

The coordinates of $A$ and $B$ are

• A. $(3,0)$ and $(0,2)$
• B. $(-\frac{8}{5},\frac{2\sqrt{161}}{15})$ and $(-\frac{9}{5},\frac{8}{5})$
• C. $(-\frac{8}{5},\frac{2\sqrt{161}}{15})$ and $(0,2)$
• D. $(3,0)$ and $(-\frac{9}{5},\frac{8}{5})$

Answer 1

The correct option is D $(3,0)$ and $(-\frac{9}{5},\frac{8}{5})$

Polar of $\frac{x^2}{9}+\frac{y^2}{4}=1$ with pole $(3,4)$ is $\frac{x}{3}+y=1$

Substituting $x=3(1-y)$ in $\frac{x^2}{9}+\frac{y^2}{4}=1$

$(y-1{)}^2+\frac{y^2}{4}=1$

Therefore, $y=0,\frac{8}{5}$

and thus $x=3,\frac{-9}{5}$