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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Parabola

Let A and B be the two points of intersection of the line y+5=0 and...

Let A and B be the two points of intersection of the line y+5=0y+5=0 and the mirror image of the parabola y2=4xy^2=4 x with respect to the line x+y+4=0x+y+4=0. If d denotes the distance between A and B , and a denotes the area of △SAB\triangle S A B, where SS is the focus of the parabola y2=4xy^2=4 x, then the value of (a+d)(a+d) is \qquad -Let A and B be the two points of intersection of the line y+5=0 and the mirror image of the parabola y^2=4 x with

Correct answer

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Step-by-step solution

Let A and B be the two points of intersection of the line y+5=0 and the mirror image of the parabola y^2=4 x with
To find image of P(t2,2t)P\left(t^2, 2 t\right)x−t21=y−2t1=−2(t2+2t+4)12+12=−(t+1)2−3x=t2−(t+1)2−3=−2t−4y=2t−(t+1)2−3=−t2−4t=−x−42⇒y+4=−(−x−42)2⇒(y+4)=−(x+4)24⇒x2=−4y⇒ Focus (−4,−5)\begin{aligned} & \frac{x-t^2}{1}=\frac{y-2 t}{1}=\frac{-2\left(t^2+2 t+4\right)}{1^2+1^2}=-(t+1)^2-3 \\ & x=t^2-(t+1)^2-3=-2 t-4 \\ & y=2 t-(t+1)^2-3=-t^2-4 \\ & t=\frac{-x-4}{2} \\ & \Rightarrow y+4=-\left(\frac{-x-4}{2}\right)^2 \\ & \Rightarrow(y+4)=-\frac{(x+4)^2}{4} \\ & \Rightarrow x^2=-4 y \\ & \Rightarrow \text{ Focus }(-4,-5) \end{aligned} Also, y=−5 intersect \text{ Also, } y=-5 \text{ intersect }∴(−4)(−1)=(x+4)24=(x+4)2x+4=±2x=−2,−6⇒d=4a=12∣101−2−51−6−51∣=12∣[1(−5+5)+1(10−30)]∣=12(20)a=10∴a+d=14\begin{aligned} & \therefore(-4)(-1)=(x+4)^2 \\ & 4=(x+4)^2 \\ & x+4= \pm 2 \\ & x=-2,-6 \\ & \Rightarrow d=4 \\ & a=\frac{1}{2}\left|\begin{array}{ccc}1 & 0 & 1 \\ -2 & -5 & 1 \\ -6 & -5 & 1\end{array}\right| \\ & =\frac{1}{2}|[1(-5+5)+1(10-30)]| \\ & =\frac{1}{2}(20) \\ & a=10 \\ & \therefore \quad a+d=14\end{aligned}$
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