Question

The locus of the point of intersection of any two perpendicular tangents to the parabola $x^2-6x+16y+41=0$ is

• A. $4y+9=0$
• B. $y-2=0$
• C. $y-3=0$
• D. $6y-13=0$

Answer 1

The correct option is B $y-2=0$

Here given parabola can be written as $(x-3{)}^2=-16(y+2)$
For a parabola $x^2=-4ay$, the locus of points of intersection of perpendicular tangents is its directrix which is $y=a$
Thus, the given locus is
$y+2=4\therefore y=2$