Question

The locus of the point of intersection of perpendicular tangents to the parabola $x^2-8x+2y+2=0$ is ?

• A. $2y-15=0$
• B. $2y+15=0$
• C. $2x+9=0$
• D. $Noneofthese$

Answer 1

Answer: (A)

$2y-15=0$

$x^2-8x+2y+2=0$

$(x^2+16-8x)+2y+2-16=0$

$(x-4{)}^2+2y-14=0$

$(x-4{)}^2+2(y-7)=0$

$(x-4{)}^2=-2(y-7)$

$(x-h{)}^2=4a(y-k)$

$4a=-2\Rightarrow a=\frac{-1}{2}$

$(h,k)=(4,7)$

$\therefore y=k-a$

$\Rightarrow y=7+\frac{1}{2}$

$\therefore 2y-15=0$