JEE Main 2025 Apr 3 Shift 2, Mathematics Q18: Involving Two or More Circles

JEE Main2025Apr 3, Shift 2Mathematics

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

A
$23 - 1$
B
$2$
C
$32 - 1$
D
$22 - 1$

Solution

Rewrite the circle: $x^{2} + (y + 6)^{2} = 1$ , centre $C = (0, - 6)$ , radius $r = 1$ .

For the parabola $y^{2} = 8 x$ (i.e. $4 a = 8$ , so $a = 2$ ), the normal at the point $(a m^{2}, - 2 am) = (2 m^{2}, - 4 m)$ has equation $y = m x - 2 am - a m^{3} = m x - 4 m - 2 m^{3}$ .

For the shortest distance between the curves, the line joining the closest points on the parabola and circle must pass through $C$ along the normal to both. So the parabola's normal passes through $C = (0, - 6)$ :

$- 6 = - 4 m - 2 m^{3} \Rightarrow m^{3} + 2 m - 3 = 0 \Rightarrow (m - 1) (m^{2} + m + 3) = 0$ .

The quadratic factor has no real roots, so $m = 1$ . The foot on the parabola is $P = (2 \cdot 1, - 4 \cdot 1) = (2, - 4)$ .

$∣ P C ∣ = (2 - 0)^{2} + (- 4 + 6)^{2} = 4 + 4 = 22$ .

Shortest distance $= ∣ P C ∣ - r = 22 - 1$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q18: Involving Two or More Circles

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