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)

$(-5,-2)$

Differentiating the equation of ellipse,
$x^2+3y^2=37$, w.r.t. $x$,
$\frac{dy}{dx}=-\frac{x}{3y}$
Slope of the normal
$m_n=-\frac{dx}{dy}=\frac{3y}{x}$
Slope of the given line is,
$\frac{3y}{x}=\frac{6}{5}\Rightarrow 2x=5y$

Hence, the points in options (b) and (d) are satisfying the above relation.