JEE 2022 — Mathematics PYQ

Step 1: Find the intersection points of the parabola and the line.

The equations are:

1. $y^2=8x$ (Parabola)

2. $y=\sqrt{2}x$ (Line)

Substitute $y=\sqrt{2}x$ into the parabola equation:

The intersection points are at $x=0$ (where $y=0$) and $x=4$ (where $y=4\sqrt{2}$).

Step 2: Calculate the total area enclosed by the parabola and the line.

The area $A_{total}$ between $x=0$ and $x=4$ is:

Step 3: Find the area of the triangle.

The triangle is formed by:

• $L_1:y=\sqrt{2}x$

• $L_2:x=1$

• $L_3:y=2\sqrt{2}$

Let's find the vertices:

1. Intersection of $y=\sqrt{2}x$ and $x=1$: $(1,\sqrt{2})$

2. Intersection of $y=2\sqrt{2}$ and $x=1$: $(1,2\sqrt{2})$

3. Intersection of $y=\sqrt{2}x$ and $y=2\sqrt{2}$: $2\sqrt{2}=\sqrt{2}x⟹x=2$. So, $(2,2\sqrt{2})$

This is a right-angled triangle with:

• Base (along $y=2\sqrt{2}$): from $x=1$ to $x=2$, so base = 1.

• Height (along $x=1$): from $y=\sqrt{2}$ to $y=2\sqrt{2}$, so height = $\sqrt{2}$.

Step 4: Subtract the triangle area from the total area.

The required area is the part of the enclosed region that lies outside the triangle. Note that the triangle defined above lies entirely within the region enclosed by the parabola and the line.

Final Answer:

The correct option is (C).