Question

A point P moves in such a way that the sum of the slopes of the normal drawn from it to the hyperbola xy = 4 is equal to the sum of the ordinates of feet of the normal. The locus of P is a parabola $x^2=4y$. Then, the least distance of this parabola from the circle $x^2-y^2-24x+128=0$ is

• A.
  
  
• B.
  
  
• C.
• D. None of the above

Answer 1

The correct option is A



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$(2t,t^2)$ be any point on the parabola $x^2=4y$ and C(12, 0) be the centre of the circle. Then,
$A{C}^2=(2t-12{)}^2+(t^2-0{)}^{2}LetZ=A{C}^2$
Then, $Z=4(t-6)2+t^4$
$\therefore$ $\frac{dz}{d{t}^2}=8(t-60+4t^3)and\frac{dz}{d{t}^2}=12+12t^2$
For maximum and minimum values of Z, we must have $\frac{dz}{d{t}^2}=0$
$\Rightarrow$ $8(t-6)+4t^3=0\Rightarrow t^3+2t-12=0$
$\Rightarrow$ $(t-2)(t^2+2t+6)=0\Rightarrow t=2$
Clearly, $\frac{dz}{d{t}^2}>0,fort=2$
Thus, Z is minimum when t = 2.
The minimum value of Z is given by $Z=4(2-6{)}^2+24=80$
$\Rightarrow$ $A{C}^2=80$ $\Rightarrow$ $AC=4\sqrt{5}$ Hence, the least distance = $4\sqrt{5}$