Question

If the shortest distance of the parabola $y^{2}=4x$ from the centre of the circle $x² + y² - 4x - 16y + 64 = 0$ is d, then d2 is equal to:

• A. 16
• B. 24
• C. 36
• D. 20

Answer 1

Answer: (D)

20

Answer 2

Step 1. Rewrite the Equation of the Circle in Standard Form

Given the equation:

$x^2 + y^2 - 4x - 16y + 64 = 0$
Completing the square for the terms involving $x$ and $y$:
$(x^2 - 4x) + (y^2 - 16y) = -64$
$(x - 2)^2 - 4 + (y - 8)^2 - 64 = -64$
Rearranging terms:
$(x - 2)^2 + (y - 8)^2 = 4$

Thus, the center of the circle is $(2, 8)$ and the radius is 2.

Step 2. Find the Normal to the Parabola

Consider the parabola $y^2 = 4x$. Let the slope of the normal be $m$. The equation of the normal to the parabola is given by:
$y = mx - 2m - m^3$
Substitute the point $(2, 8)$ into the equation to find $m$:
$8 = m \cdot 2 - 2m - m^3$
Simplifying:
$m^3 + 2m - 8 = 0$

Step 3. Calculate the Distance
The shortest distance is between the center $(2, 8)$ of the circle and the point on the parabola where the normal passes. Using the distance formula, we find:
$d^2 = (x − 2)^2 + (y − 8)^2 = 20$