Question

Find the shortest distance between the curves y² = 4x and x² + (y + 12)² = 1 

Answer 1

Given curves are

y² = 4x and x² + (y + 12)² = 1

 

Since both the tangents at P and Q are parallel.

So their slopes are same.

when y = 4, then x = 4

So the point Q is (4, 4)

Here, C = (0, – 12)

Thus, Shortest distance

= PQ

= CQ – CP

= √(16 + 64) – 1

= √80 – 1

= 2√5 – 1