Question

The point of contact $8x-9y+5=0$ with the ellipse $4x^2+9y^2=1$ is

• A. $(\frac{2}{5},\frac{1}{5})$
• B. $(-\frac{2}{5},-\frac{1}{5})$
• C. $(-\frac{2}{5},\frac{1}{5})$
• D. $(\frac{2}{5},-\frac{1}{5})$

Answer 1

Answer: (C)

$(-\frac{2}{5},\frac{1}{5})$

$4x^2+9y^2=1\Rightarrow \frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{9}}=1$

$\therefore a^2=\frac{1}{4},b^2=\frac{1}{9}$

$8x-9y+5=0\Rightarrow y=\frac{8}{9}x+\frac{5}{9}$

$\therefore m=\frac{8}{9}$

$c=\sqrt{a^2{m}^2+b^2}$

$=\sqrt{\frac{1}{4}{\left(\frac{8}{9}\right)}^2+\frac{1}{9}}$

$\therefore c=\frac{5}{9}$

P$(\frac{-a^{2}m}{c},\frac{b^2}{c})$

$=(\frac{-\frac{1}{4}\cdot \frac{8}{9}}{\frac{5}{9}},\frac{\frac{1}{9}}{\frac{5}{9}})$

$\therefore P(-\frac{2}{5},\frac{1}{5})$