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 S A B , where S is the focus of the parabola y^2=4 x , then the value of (a+d) is \underline\hspace2cm .

Parabola PYQ — JEE Main 2025 · 28 Jan · Shift 2 · Q47 with Solution

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Correct answer: 14

Given parabola and its focus

The parabola is $y^2=4x$ which is of the standard form $y^2=4ax$ with $a=1$ .

So its focus is $S=(1,0).$

Reflect the parabola across the line $x+y+4=0$

Instead of finding the reflected parabola directly, reflect the line $y+5=0$ (i.e. $y=-5$ ) across the same line. Reflection preserves incidence and distance, so the intersection of:

reflected parabola, and
original line $y=-5$

corresponds to the intersection of:

original parabola $y^2=4x$ , and
reflected line of $y=-5$ .

So we first reflect the line $y=-5$ about $x+y+4=0.$

Reflect two points on the line $y=-5$

Take two convenient points on $y=-5$ : $P_1=(0,-5), \quad P_2=(1,-5).$

For reflection of a point $(x_0,y_0)$ in the line $ax+by+c=0,$ the image is $\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right).$

Here $a=1,b=1,c=4$ .

Reflection of $P_1=(0,-5)$

We have $ax_0+by_0+c=0-5+4=-1.$ So

-5-\frac{2(1)(-1)}{2}\right)=(1,-4).$$ ### Reflection of $P_2=(1,-5)$ Now $$ax_0+by_0+c=1-5+4=0.$$ So $P_2$ lies on the mirror line, hence remains fixed: $$P_2'=(1,-5).$$ Therefore, the reflected image of the line $y=-5$ is the line through $(1,-4)$ and $(1,-5)$, i.e. $$x=1.$$ --- 4. **Intersect this reflected line with the original parabola** Now solve the intersection of $$x=1$$ and $$y^2=4x.$$ Substitute $x=1$: $$y^2=4 \implies y=\pm 2.$$ So the corresponding points on the original parabola are $$(1,2), \quad (1,-2).$$ Reflecting these back gives the required points $A,B$ on the reflected parabola cut by $y=-5$. Since reflection preserves distance, the distance between $A$ and $B$ equals the distance between these two points: $$d=\sqrt{(1-1)^2+(2-(-2))^2}=4.$$ --- 5. **Find area of $\triangle SAB$** Reflection also preserves area. So $$[\triangle SAB]=[\triangle S'A'B']$$ if all three points are reflected together. But easiest is to reflect the focus $S=(1,0)$ across the line $x+y+4=0$. Let its image be $S'$. Then $$a=[\triangle S'PQ]$$ where $$P=(1,2), \quad Q=(1,-2).$$ Now reflect $S=(1,0)$. For $(1,0)$, $$ax_0+by_0+c=1+0+4=5.$$ Thus $$S' = \left(1-\frac{2(1)(5)}{2},\;0-\frac{2(1)(5)}{2}\right)=(-4,-5).$$ So we need area of triangle with vertices $$(-4,-5),\ (1,2),\ (1,-2).$$ Segment $PQ$ is vertical with length $$PQ=4.$$ The perpendicular distance from $S'=(-4,-5)$ to the line $x=1$ is $$|{-4}-1|=5.$$ Hence area is $$a=\frac12 \times 4 \times 5=10.$$ --- 6. **Compute $a+d$** $$a+d=10+4=14.$$ --- 7. **Comparison with stored answer** Derived answer is $$14$$ which matches the stored correct answer.

More from Parabola

View written solutionFree

Correct answer: 14

Given parabola and its focus

The parabola is $y^2=4x$ which is of the standard form $y^2=4ax$ with $a=1$ .

So its focus is $S=(1,0).$

Reflect the parabola across the line $x+y+4=0$

Instead of finding the reflected parabola directly, reflect the line $y+5=0$ (i.e. $y=-5$ ) across the same line. Reflection preserves incidence and distance, so the intersection of:

reflected parabola, and
original line $y=-5$

corresponds to the intersection of:

original parabola $y^2=4x$ , and
reflected line of $y=-5$ .

So we first reflect the line $y=-5$ about $x+y+4=0.$

Reflect two points on the line $y=-5$

Take two convenient points on $y=-5$ : $P_1=(0,-5), \quad P_2=(1,-5).$

For reflection of a point $(x_0,y_0)$ in the line $ax+by+c=0,$ the image is $\left(x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\; y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}\right).$

Here $a=1,b=1,c=4$ .

Reflection of $P_1=(0,-5)$

We have $ax_0+by_0+c=0-5+4=-1.$ So

-5-\frac{2(1)(-1)}{2}\right)=(1,-4).$$ ### Reflection of $P_2=(1,-5)$ Now $$ax_0+by_0+c=1-5+4=0.$$ So $P_2$ lies on the mirror line, hence remains fixed: $$P_2'=(1,-5).$$ Therefore, the reflected image of the line $y=-5$ is the line through $(1,-4)$ and $(1,-5)$, i.e. $$x=1.$$ --- 4. **Intersect this reflected line with the original parabola** Now solve the intersection of $$x=1$$ and $$y^2=4x.$$ Substitute $x=1$: $$y^2=4 \implies y=\pm 2.$$ So the corresponding points on the original parabola are $$(1,2), \quad (1,-2).$$ Reflecting these back gives the required points $A,B$ on the reflected parabola cut by $y=-5$. Since reflection preserves distance, the distance between $A$ and $B$ equals the distance between these two points: $$d=\sqrt{(1-1)^2+(2-(-2))^2}=4.$$ --- 5. **Find area of $\triangle SAB$** Reflection also preserves area. So $$[\triangle SAB]=[\triangle S'A'B']$$ if all three points are reflected together. But easiest is to reflect the focus $S=(1,0)$ across the line $x+y+4=0$. Let its image be $S'$. Then $$a=[\triangle S'PQ]$$ where $$P=(1,2), \quad Q=(1,-2).$$ Now reflect $S=(1,0)$. For $(1,0)$, $$ax_0+by_0+c=1+0+4=5.$$ Thus $$S' = \left(1-\frac{2(1)(5)}{2},\;0-\frac{2(1)(5)}{2}\right)=(-4,-5).$$ So we need area of triangle with vertices $$(-4,-5),\ (1,2),\ (1,-2).$$ Segment $PQ$ is vertical with length $$PQ=4.$$ The perpendicular distance from $S'=(-4,-5)$ to the line $x=1$ is $$|{-4}-1|=5.$$ Hence area is $$a=\frac12 \times 4 \times 5=10.$$ --- 6. **Compute $a+d$** $$a+d=10+4=14.$$ --- 7. **Comparison with stored answer** Derived answer is $$14$$ which matches the stored correct answer.

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