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 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 $(-\frac{9}{5},\frac{8}{5})$

Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$ ..... $\left(i\right)$

Equation of chord $AB$ is

$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$

$\frac{3x}{9}+\frac{4y}{4}=1$

$x+3y-3=0$

$x=3(-y+1)$

Now, from $\left(i\right)$

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

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

$\frac{5y^2}{4}-2y=0$

$y(\frac{5y}{4}-2)=0$

$y=0$ or $\frac{8}{5}$

$\Rightarrow x=3-3y$

$x=3$ when $y=0$

$x=3-3\times \frac{8}{5}$ when $y=\frac{8}{5}$

$x\Rightarrow -\frac{9}{5}$

$(x,y)=(3,0)$ or $(\frac{-9}{5},\frac{8}{5})$