Question

The vertex of the conic represented by $25(x^2+y^2)=(3x-4y+12{)}^2$ is

• A. $(\frac{24}{25},-\frac{18}{25})$
• B. $(0,0)$
• C. $(-\frac{36}{25},\frac{48}{25})$
• D. $(-\frac{18}{25},\frac{24}{25})$

Answer 1

The correct option is D $(-\frac{18}{25},\frac{24}{25})$

Given conic is $25(x^2+y^2)=(3x-4y+12{)}^2$
$\Rightarrow \sqrt{x^2+y^2}=\left|\frac{3x-4y+12}{5}\right|$
So distance of $(x,y)$ from the origin is same as the distance from origin to the line $3x-4y+12=0$
This is locus of a parabola with focus $=(0,0)$ and equation of directrix $3x-4y+12=0$

Let the foot of perpendicular to the directrix from focus be $(h,k)$. Then
$\frac{h-0}{3}=\frac{k-0}{-4}=-\frac{12}{3^2+4^2}\Rightarrow h=-\frac{36}{25},k=\frac{48}{25}$

We know that the vertex is the midpoint of focus and foot of perpendicular.
So the vertex is $(\frac{h}{2},\frac{k}{2})=(-\frac{18}{25},\frac{24}{25})$