Exam

JEE Main

Year

2025

Date

2 April

Shift

Evening Shift

Subject

Mathematics

Chapter

Parabola

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Parabola

Let the point P of the focal chord P Q of the parabola y^2=16 x be...

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 PQ in the ratio $\mathrm{m}: \mathrm{n}$, $\operatorname{gcd}(m, n)=1$, then $m^2+n^2$ is equal to :

Step-by-step solution

$y^2=16 x ; a=4 \quad$ focus $S \equiv(4,0)$

$$\begin{aligned} & 2 \mathrm{at}_1=-4 \\ & \Rightarrow 2(4) \mathrm{t}_1=-4 \\ & \Rightarrow \mathrm{t}_1=-\frac{1}{2} \\ & \because \mathrm{t}_1 \mathrm{t}_2=-1\end{aligned}$$
$$\begin{aligned}<br/>& \Rightarrow \mathrm{t}_2=2 \\ & \therefore \mathrm{Q}\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)=(16,16)<br/>\end{aligned}$$
Let, S divides PQ internally in $\lambda: 1$ ratio
$$\begin{aligned}<br/>& \therefore \frac{16 \lambda-4}{\lambda+1}=0 \\ & \lambda=\frac{1}{4}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \therefore \mathrm{~m}^2+\mathrm{n}^2=1+16=17<br/>\end{aligned}$$