Mathematics · Parabola

JEE Main 2025 — 28 January, Evening Shift — Question 24

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 respect to the line $x+y+4=0$ . If $d$ denotes the distance between A and B , and a denotes the area of $\triangle \mathrm{SAB}$ , where S is the focus of the parabola $y^{2}=4 x$ , then the value of $(a+d)$ is

Answer: 14

Numerical answer — enter this value.

Step-by-step solution

\begin{aligned} &\text{Given\; parabola\; } y^2=4x\; \text{ and\; line\; of\; reflection\; } L: x+y+4=0.\\[4pt] &\text{Reflection\; formula\; across\; } ax+by+c=0\ (a=b=1,c=4):\\ &\qquad (x',y')=\Big(x-\frac{2a(ax+by+c)}{a^2+b^2},\;y-\frac{2b(ax+by+c)}{a^2+b^2}\Big)\\[4pt] &\text{Hence\; for\; }L:\quad (x',y')=(x-(x+y+4),\;y-(x+y+4))=(-y-4,\,-x-4).\\[6pt] &\text{We\; want\; the\; image\; point\; }(x',y')\; \text{ to\; lie\; on\; }y'=-5.\\ &\qquad -x-4=-5\implies x=1.\\[4pt] &\text{Points\; on\; the\; parabola\; with\; }x=1\text{ satisfy\; }y^2=4\cdot1=4\implies y=\pm2.\\[4pt] &\text{Their\; reflections\; are} \begin{aligned} &(1,2)\mapsto(-2-4,\,-1-4)=(-6,-5),\\ &(1,-2)\mapsto(2-4,\,-1-4)=(-2,-5). \end{aligned}\\[6pt] &\text{Thus\; }A=(-6,-5),\ B=(-2,-5).\\[6pt] &d=|{-2}-(-6)|=4.\\[6pt] &\text{Focus\; of \;}y^2=4x\; \text{ is\; }S=(1,0). \text{ The\; altitude\; from\; }S\text{ to\; base\; }AB\ (y=-5)\text{ is\; }5.\\[4pt] &\text{Area\; }a=\tfrac{1}{2}\cdot(\text{base\; }d)\cdot(\text{height\; }5)=\tfrac{1}{2}\cdot4\cdot5=10.\\[6pt] &\therefore\quad a+d=10+4=\boxed{14}. \end{aligned}

Answer key and solution verified before publishing.

Practise Parabola

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Parabola
Topic: Pole and Polar wrt a Parabola