Let A and B be the two points of intersection of the line y + 5 = 0 and the mirror image of the parabola with respect to the line x + y + 4 = 0. If d denotes the distance between A and B, and a denotes the area of SAB, where S is the focus of the parabol

Question

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 $△$ SAB, where S is the focus of the parabola $y^{2} = 4 x$ , then the value of (a + d) is __________. [2025]

Smart Achievers · Accepted Answer

(14)

Image of point (0, 0) w.r.t. to line x + y + 4 = 0

$\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = - 2 \frac{(a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$

$\frac{x - 0}{1} = \frac{y - 0}{1} = \frac{- 2 (4)}{2} \Rightarrow x = y = - 4$

Image of focus (1, 0) w.r.t. to line x + y + 4 = 0

$\frac{x - 1}{1} = \frac{y - 0}{1} = - 2 \frac{(1 + 4)}{2} x = - 4; y = - 5$

Equation of mirror image of parabola

${(x - h)}^{2} = - 4 a (y - k) {(x + 4)}^{2} = - 4 (1) (y + 4)$

Put y = –5; we get x = –6 and –2

$∴$ A = (–6, –5); B = (–2, –5)

Distance between the points, d = AB = 4

Area of $△$ SAB, $a = \frac{1}{2} \times 4 \times 5 = 10$

So, a + d = 14.

Solution

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