Question

Coordinates of point(s) on the ellipse $x^2+3y^2=37,$ where the normal is parallel to the line $6x-5y=2,$ is/are

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

Answer 1

Answer: (B), (D)

$(-5,-2)$

Equation of the ellipse is:
$\frac{x^2}{37}+\frac{y^2}{\left(\frac{37}{3}\right)}=1$

The equation of the normal at point $P(\sqrt{37}\cos \theta ,\sqrt{\frac{37}{3}}\sin \theta )$ is :
$\left(\sqrt{37}\sec \theta \right)x-\left(\sqrt{\frac{37}{3}}\text{cosec}\theta \right)y=37-\frac{37}{3}$
$\Rightarrow \left(\sqrt{37}\sec \theta \right)x-\left(\sqrt{\frac{37}{3}}\text{cosec}\theta \right)y=\frac{74}{3}\dots \left(1\right)$

If the normal is parallel to the line $6x-5y=2,$ then
$\Rightarrow \frac{6}{5}=\sqrt{3}\tan \theta$
$\Rightarrow \cos \theta =\frac{5}{\sqrt{37}},\sin \theta =\frac{2\sqrt{3}}{\sqrt{37}}$
or
$\Rightarrow \cos \theta =-\frac{5}{\sqrt{37}},\sin \theta =-\frac{2\sqrt{3}}{\sqrt{37}}$

Hence, point $P$ can be $(5,2)$ or $(-5,-2).$

Alternate Solution:
Equation of normal on $x^2+3y^2=37$ will be
$y=mx\pm \frac{(a^2-b^2)m}{\sqrt{a^2+b^2{m}^2}}$
$\because$ Normal is parallel to $6x-5y=2$
$\therefore m=\frac{6}{5}$
Hence, normal will be
$5y=6x\pm \frac{30(37-37/3)}{37}\Rightarrow \frac{37x}{\pm 5}-\frac{\left(37/3\right)\times y}{\pm 2}=\frac{74}{3}$
Equation of normal
$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
Hence, point of contact will be
$(5,2)$ and $(-5,-2)$