Question

The locus of the point which moves, so that its distance from the fixed point $(-2,3)$ equals its distance from the line $x+6=0$, is

• A. $y^2-6y+8x=0$
• B. $y^2-3y+2x+5=0$
• C. $y^2-3y+2x-5=0$
• D. $y^2-6y+8x+23=0$
• E. $y^2-6y-8x-23=0$

Answer 1

Answer: (E)

$y^2-6y-8x-23=0$

Let the point is $P(h,k)$.

By definition of parabola $a$.

The locus of point $P(h,k)$ which moves in a plane such that its distance from a fixed point $(-2,3)S$ is always in a constant ratio to its perpendicular distance from a fixed straight line $(x+6=0)M$.
i.e., $\frac{PS}{PM}=1(\because e=1$ for parabola)
$\Rightarrow (PS{)}^2=(PM{)}^2$
$\Rightarrow (h+2{)}^2+(k-3{)}^2=\frac{|h+6{|}^2}{(\sqrt{(1{)}^2}{)}^2}$
$\Rightarrow h^2+4+4h+k^2+9-6k=(h+6{)}^2$
$\Rightarrow h^2+k^2+4h-6k+13=h^2+36+12h$
$\Rightarrow k^2-8h-6k-23=0$
Required locus is $y^2-8x-6y-23=0$.