JEE Mathematics 2025 Question with Solution

Given: A line through $A(-2,0)$ touches the parabola $y^2=x-2$ at a point $B$ in the first quadrant.

Find: The area bounded by the line $AB$, the parabola, and the $x$-axis.

Using the tangent form through $A(-2,0)$,

$$y=m(x+2)$$

Substituting into the parabola,

$$y^2=x-2$$

gives

$$(m(x+2))^2=x-2$$

which leads to the quadratic equation

$$m^2x^2+(4m^2-1)x+(4m^2+2)=0$$

For tangency, the discriminant must be zero:

$$(4m^2-1)^2-4m^2(4m^2+2)=0$$

Solving, we get

$$m=\frac{1}{4}$$

Hence the tangent is

$$y=\frac{1}{4}(x+2)$$

or equivalently

$$x=4y-2$$

The point of tangency is $B=(6,2)$.

Now the required area is the area between $x=y^2+2$ and $x=4y-2$ from $y=0$ to $y=2$:

$$A=\int_0^2 \left((y^2+2)-(4y-2)\right) \, dy$$ $$A=\int_0^2 (y^2-4y+4) \, dy$$ $$A=\left[\frac{(y-2)^3}{3}\right]_0^2=\frac{8}{3}$$

Therefore, the area of the required region is $\frac{8}{3}$. The correct option is C.

Given: The parabola is $y^2=x-2$ and the tangent passes through $A(-2,0)$.

Find: The enclosed area.

Write the parabola as $y^2=4a(x-2)$, so $4a=1$ and hence $a=\frac{1}{4}$. A parametric point on the parabola is

$$(t^2+2,t)$$

The slope of the tangent there is

$$m_T=\frac{1}{2t}$$

Since the tangent passes through $A(-2,0)$ and the point $\left(t^2+2,t\right)$, its slope is

$$\frac{t-0}{(t^2+2)-(-2)}=\frac{t}{t^2+4}$$

Equating slopes,

$$\frac{t}{t^2+4}=\frac{1}{2t}$$

so

$$2t^2=t^2+4$$ $$t^2=4$$

As the point is in the first quadrant, $t=2$. Thus $B=(6,2)$ and the tangent is

$$x-4y+2=0$$

that is,

$$x=4y-2$$

Now compute the area:

$$A=\int_0^2 \left((y^2+2)-(4y-2)\right) \, dy=\frac{8}{3}$$

Therefore, the required area is $\frac{8}{3}$.