If the shortest distance of the parabola y²=4 x from the centre of the circle x²+y²-4 x-16 y+64=0 is d, then d² is equal to :

Solution

Maths sir · 4 steps

The shortest line from a point to a parabola meets it along a normal, so find the normal through the centre.

1. Find the centre. Compare x²+y²-4x-16y+64=0 with x²+y²+2gx+2fy+c=0. The centre is (-g,-f)=(2,8).
2. Use the normal. The closest point P on the parabola is where the segment from (2,8) is perpendicular to the curve, so it lies along the normal at P. For y²=4x take P=(t²,2t). The normal at P is:

   y = -tx + 2t + t³

3. Pass through centre. Put (2,8) into the normal: 8=-2t+2t+t³, so t³=8 and t=2. The foot of the normal is P=(4,4).
4. So. d²=(4-2)²+(4-8)²=4+16= 20, which is option (c).

The idea to keep. For the shortest distance from a point to a curve, draw the normal from that point to the curve. Use the parametric normal, make it pass through the point, and solve for t.

Answer from the NTA final answer key. Solution by PrepWiser.