Question

The length of the perpendicular from the origin, on the normal to the curve, $x^2+2xy-3y^2=0$ at the point $(2,2)$ is :

• A. $2$
• B. $2\sqrt{2}$
• C. $4\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$x^2+2xy-3y^2=0$
$m_N=$ slope of normal drawn to curve at $\left(2,2\right)$ is $-1$
$L:x+y=4$.
perpendicular distance of L from $\left(0,0\right)$
$=\frac{|0+0-4|}{\sqrt{2}}=2\sqrt{2}$