First slide

Maxima and minima

Question

$The coordinates of the point on the parabola y^{2} = 8 x which is at minimum distance from the circle x^{2} + (y + 6)^{2} = 1 are$

Moderate

A

$(2, - 4)$
B

$(18, - 12)$
C

$(2, 4)$
D

$None of these$

Solution

Let $P (2 t^{2}, 4 t)$ be any point on the parabola. The centre of the given circle is O (0, -6) and the radius is 1
$\begin{array}{l} O P^{2} = 4 t^{4} + (4 t + 6)^{2} = 4 [t^{4} + 4 t^{2} + 9 + 12 t] \\ = 4 x, where x = t^{4} + 4 t^{2} + 12 t + 9 \end{array}$
$\begin{array}{l} \frac{d x}{d t} = 4 t^{3} + 8 t + 12 = 4 (t^{3} + 2 t + 3) \\ = 4 (t + 1) (t^{2} - t + 3) \\ So \frac{d x}{d t} = 0 \Rightarrow t = - 1 \\ (other roots are imaginary) \\ So \frac{d^{2} x}{d t^{2}} = 4 (3 t^{2} + 2), {\frac{d^{2} x}{d t^{2}}|}_{t = - 1} > 0 \end{array}$
$Hence O P^{2} is minimum at t = - 1 . But if A is any point on$ $the circle and on OP (min), then AP will be minimum when$
$OP is minimum as AP = OP - (radius of circle), Thus the required point is P (2 (- 1))^{2}, 4 (- 1)) = (2, - 4)$

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