JEE Main 2025 Apr 2 Shift 2, Mathematics Q17: Parabola: Chord, Tangent and Normal

JEE Main2025Apr 2, Shift 2Mathematics

Let the point $P$ of the focal chord $P Q$ of the parabola $y^{2} = 16 x$ be $(1, - 4)$ . If the focus of the parabola divides the chord $P Q$ in the ratio $m : n$ , $g cd (m, n) = 1$ , then $m^{2} + n^{2}$ is equal to :

A
$17$
B
$10$
C
$37$
D
$26$

Solution

The parabola $y^{2} = 16 x$ has $4 a = 16$ , so $a = 4$ and focus $S = (4, 0)$ .

Parametric form: a point is $(a t^{2}, 2 a t) = (4 t^{2}, 8 t)$ . For $P (1, - 4)$ : $8 t_{1} = - 4 \Rightarrow t_{1} = - \frac{1}{2}$ .

For a focal chord $t_{1} t_{2} = - 1$ , so $t_{2} = 2$ and $Q = (4 \cdot 4, 8 \cdot 2) = (16, 16)$ .

The focus divides $P Q$ in ratio $m : n$ where $m = S P$ (focal distance to $P$ ) and $n = S Q$ (to $Q$ ). Using $r = a (1 + t^{2})$ : $S P = 4 (1 + \frac{1}{4}) = 5$ and $S Q = 4 (1 + 4) = 20$ . So $m : n = 5 : 20 = 1 : 4$ .

With $g cd (m, n) = 1$ , $m = 1, n = 4 \Rightarrow m^{2} + n^{2} = 1 + 16 = 17$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q17: Parabola: Chord, Tangent and Normal

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