Question

A point P moves in such a way that the sum of the slopes of the normals drawn from itto the hyperbola xy = 4 is equal to the sum of the ordinates of feet of the normals. The locus of P is a parabola x² = 4y. Then the least distance of this parabola from the circle x² – y² – 24x + 128 = 0 is –

• A. 4$\sqrt{5}$
• B. $\frac{4}{\sqrt{5}}$
• C. –$4\sqrt{5}$
• D. None of these

Answer 1

Answer: (A)

4$\sqrt{5}$

Answer 2

The distance of the parabola from the circle means the distance of any point on the parabola from the centre of the circle.

Let A(2t, t²) be any point on the parabola x² = 4y and C(12, 0) be the centre of the circle. Then, AC² = (2t – 12)² + (t² – 0)²

Let Z = AC². Then,

Z = 4(t – 6)² + t⁴

\ $\frac{dZ}{dt}$= 8 (t – 6) + 4t³ and $\frac{d^{2}Z}{dt^{2}}$ = 8 + 12t²

For maximum and minimum values of Z, we must have$\frac{dZ}{dt} = 0$

Ž 8(t – 6) + 4t³ = 0

Ž t³ + 2t – 12 = 0

Ž (t – 2) (t² + 2t + 6) = 0

Ž t = 2

Clearly, $\frac{d^{2}Z}{dt^{2}}$ > 0 for t = 2.

Thus Z is minimum when t = 2. The minimum value of Z is given by Z = 4 (2 – 6)² + 2⁴ = 80

Ž AC² = 80 Ž AC = 4$\sqrt{5}$

Hence, the least distance = 4$\sqrt{5}$.