The shortest distance between the curves and is: [2025]

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Solution

Q.
The shortest distance between the curves $y^{2} = 8 x$ and $x^{2} + y^{2} + 12 y + 35 = 0$ is: [2025]

1 $3 \sqrt{2} - 1$

2 $2 \sqrt{3} - 1$

3 $\sqrt{2}$

4 $2 \sqrt{2} - 1$

Ans.
(4)

Given, $x^{2} + y^{2} + 12 y + 35 = 0$ , which is a circle having centre, C = (0, –6)

Radius, $r = \sqrt{36 - 35} = 1$

Since, the shortest distance between both curves will be normal to the curve $y^{2} = 8 x$ from centre (0, –6), as shown in figure.

$∴$ Equation of normal to the parabola

$y^{2} = 8 x$ is $y = m x - 4 m - 2 m^{3}$

where m is slope, passes through (0, –6), then –6 = –4m – $2 m^{3}$

$\Rightarrow m^{3} + 2 m - 3 = 0$

$\Rightarrow (m - 1) (m^{2} + m + 3) = 0$

$\Rightarrow m = 1$

$P \equiv (a m^{2}, - 2 a m) = (2, - 4)$

$∴$ Shortest distance, PQ = PC – r

$= \sqrt{{(2 - 0)}^{2} + {(- 4 + 6)}^{2}} - 1 = \sqrt{8} - 1 = (2 \sqrt{2} - 1) units$ .

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