Question

A point on the ellipse $x^2+3y^2=37$, where the normal is parallel to the line $6x-5y=2$ is

• A. $(5,-2)$
• B. $(5,2)$
• C. $(-5,2)$
• D. $(-5,-2)$

Answer 1

Answer: (B), (D)

The correct options are
B $(5,2)$
D $(-5,-2)$

Given ellipse may be written as, $\frac{x^2}{37}+\frac{y^2}{37/3}=1$

$\Rightarrow a^2=37,b^2=37/3$

Thus general equation of normal to ellipse with slope 'm' is given by,

$y=mx-\frac{(a^2-b^2)m}{\sqrt{a^2+b^2{m}^2}}=mx\pm \frac{74/3m}{\sqrt{37+37/3m^2}}$

we have $m=6/5$

$\Rightarrow y=6/5x\pm \frac{74/3\times 6/5}{\sqrt{37+37/3\times 36/25}}$

$\Rightarrow y=6/5x\pm 4$

$\Rightarrow 6x-5y\pm 20=0..\left(1\right)$

Solving line (1) with the given ellipse we get point of intersections which are, $(5,2)$ and $(-5,-2).$