Question

A point on the ellipse $4x^2+9y^2=36$, where the normal is parallel to the line, $4x-2y-5=0$, is:

• A. $(\frac{8}{5},\frac{9}{5})$
• B. $(-\frac{9}{5},\frac{8}{5})$
• C. $(\frac{8}{5},\frac{9}{4})$
• D. $(\frac{9}{5},\frac{8}{5})$

Answer 1

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

Point : $(3cos\theta ,2sin\theta )$

Normal $3xsec\theta -2ycosec\theta =9-4$

$=5$

$2ycosec\theta =3xsec\theta -5$

Slope = $\frac{3sec\theta }{2cossec\theta }=\frac{4}{2}$

$\Rightarrow tan\theta =\frac{4}{3}$

$\therefore cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

$\therefore$ point $=\frac{9}{5},\frac{8}{5}$