Let the point P of the focal chord PQ of the parabola be (1, –4). If the focus of the parabola divides the chord PQ in the ration m : n, gcd (m, n) = 1, then is equal to : [2025]

Question

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 : [2025]

Smart Achievers · Accepted Answer

(2)

End point of the focal chord PQ of the parabola $y^{2} = 4 a x$ is $P (a t^{2}, 2 a t)$

Now, we have parabola $y^{2} = 16 x$

$∴ P (4 t^{2}, 8 t) = (1, - 4)$

$\Rightarrow t = \frac{- 1}{2}$

$∴$ Point Q is given by $(\frac{a}{t^{2}}, - \frac{2 a}{t}) = (16, 16)$

Now, focus of parabola $y^{2} = 16 x$ is (4, 0)

Focus (4, 0) divides P(1, –4) and Q(16, 16) in ratio m : n

$\Rightarrow (4, 0) = (\frac{16 m + n}{m + n}, \frac{16 m - 4 n}{m + n})$

$\Rightarrow 16 m - 4 n = 0 \Rightarrow 4 m = n \Rightarrow \frac{m}{n} = \frac{1}{4}$

$\Rightarrow m = 1, n = 4$ [ $∵$ gcd(m, n) = 1]

$∴ m^{2} + n^{2} = 1 + 16 = 17$ .

Solution

Q.
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 : [2025]

1 10

2 17

3 26

4 37

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