First slide

Parabola

Question

$The minimum distance between the curves y^{2} = 4 x and x^{2} + y^{2} - 12 x + 31 = 0 is$

Moderate

A

$\sqrt{2}$
B

$\sqrt{5}$
C

$2 \sqrt{5}$
D

$\sqrt{10}$

Solution

$Centre and radius of the given circle are P (6, 0) and \sqrt{5}, respectively.$
Now, the shortest distance always occurs along the common normal.
$Differentiating y^{2} = 4 x with respect to x we get,$
$\frac{d y}{d x} = \frac{2}{y}$
$Then the slope of nonnal at point A (y_{1}^{2} / 4, y_{1}) is - y_{1} / 2$
$Also, from definition, the slope of A P is given by \frac{y_{1} - 0}{(y_{1}^{2} / 4) - 6} = - \frac{y_{1}}{2}$
$i.e., y_{1} = 0 or y_{1} = \pm 4$
$Hence, the points are O (0, 0), A (4, 4), and C (4, - 4) .$
$The shortest distance is A P - \sqrt{5} = \sqrt{20} - \sqrt{5} = \sqrt{5} .$

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