Let be the point on the parabola which is at the shortest distance from the center of the circle Let be the point on the circle dividing the line segment internally. Then [2016]

Question

Let $P$ be the point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0 .$ Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then [2016]

Smart Achievers · Accepted Answer

(1, 3, 4)

Let point $y^{2} = 4 x$ on parabola $y^{2} = 4 x$ be $x^{2} + y^{2} - 4 x - 16 y + 64 = 0 .$

$∵$ PS is shortest distance, therefore PS should be the normal to parabola.

Equation of normal to $y^{2} = 4 x$ at $P (t^{2}, 2 t)$ is

$y - 2 t = - t (x - t^{2})$

It passes through $S (2, 8)$

$∴$ $8 - 2 t = - t (2 - t^{2})$

$\Rightarrow t^{3} = 8 \Rightarrow t = 2,$ $∴$ $P (4, 4)$

Also slope of tangent to circle at $Q = \frac{- 1}{slope of P S} = \frac{1}{2}$

Equation of normal at $t = 2$ is $2 x + y = 12$

Clearly $x$ -intercept = 6, Now $S P = 2 \sqrt{5}$ and $S Q = r = 2$

$∴$ $Q divides S P in the ratio S P : P Q$

$= 2 : 2 (\sqrt{5} - 1) or (\sqrt{5} + 1) : 4$

Solution

Q.
Let $P$ be the point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0 .$ Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then [2016]

1 $S P = 2 \sqrt{5}$

2 $S Q : Q P = (\sqrt{5} + 1) : 2$

3 the $x$ -intercept of the normal to the parabola at $P$ is 6

4 the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let P be the point on the parabola y2=4x which is at the shortest distance from the center S of the circle x2+y2-4x-16y+64=0. Let Q be the point on the circle dividing the line segment SP internally. Then [2016]

1 SP=25

2 SQ:QP=(5+1):2

3 the x-intercept of the normal to the parabola at P is 6