Question ID: #500
Let A and B be the two points of intersection of the line $y+5=0$ and the mirror image of the parabola $y^{2}=4x$ with respect to the line $x+y+4=0.$ If $d$ denotes the distance between A and B, and $a$ denotes the area of $\Delta SAB$, where S is the focus of the parabola $y^{2}=4x,$ then the value of $(a+d)$ is:
Solution:
Original Parabola: $y^2 = 4x$.
Focus $S(1, 0)$. Vertex $V(0, 0)$.
The mirror line is $x + y + 4 = 0$.
Let’s find the image of the focus $S(1, 0)$ with respect to the line. Let image be $S'(h, k)$.
$$\frac{h-1}{1} = \frac{k-0}{1} = -2 \frac{1(1) + 1(0) + 4}{1^2 + 1^2}$$
$$\frac{h-1}{1} = \frac{k}{1} = -2 \frac{5}{2} = -5$$
$$h = -4, k = -5 \Rightarrow S'(-4, -5)$$
Now find image of Vertex $V(0,0)$. Let image be $V'(x_1, y_1)$.
$$\frac{x_1-0}{1} = \frac{y_1-0}{1} = -2 \frac{4}{2} = -4$$
$$V'(-4, -4)$$
The axis of the original parabola is $y=0$ (Horizontal). The mirror line is at $135^\circ$. The reflected axis will be vertical.
Since $V'(-4, -4)$ and $S'(-4, -5)$, the new axis is $x = -4$.
The parabola opens downwards (since $S’$ is below $V’$).
Distance $a’ = V’S’ = 1$.
Equation of reflected parabola: $(x – (-4))^2 = -4a'(y – (-4))$.
$$(x+4)^2 = -4(y+4)$$
We need intersection with line $y + 5 = 0 \Rightarrow y = -5$.
$$(x+4)^2 = -4(-5+4) = -4(-1) = 4$$
$$x+4 = \pm 2$$
$$x = -2 \text{ or } x = -6$$
Points are $A(-6, -5)$ and $B(-2, -5)$.
Distance $d = AB = |-2 – (-6)| = 4$.
Now, Area of $\Delta SAB$. $S(1, 0)$.
Base $AB$ is on line $y = -5$. Length $= 4$.
Height of $\Delta SAB$ is perpendicular distance from $S(1, 0)$ to $y = -5$.
$$Height = |0 – (-5)| = 5$$
$$Area (a) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10$$
Value of $a + d = 10 + 4 = 14$.
Ans. (14)
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