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

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Parabola

Let the point P of the focal chord PQ of the parabola

y^{2} = 16 x

(1, - 4)

. If the focus of the parabola divides the chord PQ in the ratio

m : n

g cd (m, n) = 1

, then

m^{2} + n^{2}

is equal to :

(A) $17$ ✓ Correct
(B) $10$
(C) $37$
(D) $26$

Answer: (A)

Solution

For

y^{2} = 16 x

4 a = 16

a = 4

and the focus is

(4, 0)

. Using the parametric point

(4 t^{2}, 8 t)

, the point

P (1, - 4)

gives

8 t_{1} = - 4

, so

t_{1} = - \frac{1}{2}

. For a focal chord

t_{1} t_{2} = - 1

, hence

t_{2} = 2

and

Q = (16, 16)

. If the focus divides

P Q

in ratio

m : n

, then

\frac{16 m + n}{m + n} = 4

gives

3 n = 12 m

, i.e.

m : n = 1 : 4

. With

g cd (1, 4) = 1

m^{2} + n^{2} = 1 + 16 = 17

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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