Question

On the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$, one of the points at which the normals are parallel to the line $2x-y=1$ is

• A. $(\frac{9}{\sqrt{10}},\frac{2}{\sqrt{10}})$
• B. $(-\frac{9}{\sqrt{10}},\frac{2}{\sqrt{10}})$
• C. $(\frac{2}{\sqrt{10}},\frac{9}{\sqrt{10}})$
• D. None of these

Answer 1

The correct option is C $(\frac{2}{\sqrt{10}},\frac{9}{\sqrt{10}})$

Given equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{9}=1$

Let the feet of normal be $P(x,y)$

Normal is parallel to $2x-y=1$

Therefore, slope of normal at $P$ $=2$

and slope of slope of tangent at $P$ $=-\frac{1}{2}$

Point of contact of tangent in slope from is $(\frac{\pm a^{2}m}{\sqrt{a^2{m}^2+b^2}},\frac{{\mp b}^2}{\sqrt{a^2{m}^2+b^2}})$

Here $a=2,b=3$ and $m=-\frac{1}{2}$

Therefore, the point of contact are:

$(\frac{\pm (4\times \frac{-1}{2})}{\sqrt{4\times \frac{1}{4}+9}},\frac{\mp 9}{\sqrt{4\times \frac{1}{4}+9}})(\frac{\mp 2}{\sqrt{10}},\frac{\mp 9}{\sqrt{10}})$

So, option C is correct.