Let the point $P$ of the focal chord $PQ$ of the parabola ${y}^{2}=16x$ be $(1,-4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m:n$, $\gcd (\mathrm{m},\mathrm{n})=1$, then ${m}^{2}+{n}^{2}$ is equal to : [JEE Main 2025, 2 Apr (Shift 2)]

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Correct answer: a) $17$. $P(a{t}^{2},2at)$ $\Rightarrow P(4{t}^{2},8t)$ $=(1,-4)$ $\Rightarrow t=\frac{−1}{2}$;$Q\left(\frac{a}{{t}^{2}},\frac{−2a}{t}\right)$ $S(4,0)$ is the focus & $PS=a+{a}^{2}$ $QS=a+\frac{a}{{t}^{2}}$ $\frac{PS}{QS}={t}^{2}=\frac{4}{1}=\frac{{m}^{}}{n}$ ${m}^{2}+{n}^{2}=17$

Let the point \(P\) of the focal chord \(PQ\) of the parabola \({y}^{2}=16x\) be \((1,-4)\). If the focus of the parabol…

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