JEE Mathematics 2023 Question with Solution

Given: The curves are $y^2 - 2y = -x$ and $x + y = 0$.

Find: The value of $8A$, where $A$ is the area bounded by the two curves.

From $x + y = 0$, we get

$$x = -y$$

Also, from $y^2 - 2y = -x$,

$$x = -y^2 + 2y$$

Now substitute $x = -y$ into the first curve:

$$y^2 - 2y = -(-y)$$ $$y^2 - 3y = 0$$ $$y(y-3)=0$$

So the points of intersection correspond to

$$y=0, 3$$

Hence the intersection points are $(0,0)$ and $(-3,3)$.

The area between the curves, integrating with respect to $y$, is

$$A = \int_{0}^{3} \left[(-y) - (-y^2 + 2y)\right] \, dy$$ $$A = \int_{0}^{3} (y^2 - 3y) \, dy$$

Evaluating,

$$A = \left[\frac{y^3}{3} - \frac{3y^2}{2}\right]_{0}^{3}$$ $$A = 9 - \frac{27}{2} = \frac{9}{2}$$

Therefore,

$$8A = 8 \times \frac{9}{2} = 36$$

So, the required answer is $36$.

Given: $y^2 - 2y = -x$ and $x + y = 0$.

Find: The value of $8A$.

Rewrite the equations in terms of $x$:

$$x = -y^2 + 2y$$ $$x = -y$$

Substitute the line into the parabola to find intersection points:

$$y^2 - 2y = y$$ $$y^2 - 3y = 0$$ $$y(y-3)=0$$

Thus,

$$y=0 \text{ or } y=3$$

Using $x=-y$, the points are

$$(0,0) \text{ and } (-3,3)$$

Now the horizontal distance between the curves is taken from the parabola to the line:

$$A = \int_{0}^{3} \left[(-y^2+2y)-(-y)\right] \, dy$$ $$A = \int_{0}^{3} (-y^2+3y) \, dy$$ $$A = \left[-\frac{y^3}{3} + \frac{3y^2}{2}\right]_{0}^{3}$$ $$A = -9 + \frac{27}{2} = \frac{9}{2}$$

Therefore,

$$8A = 36$$

The correct numerical value is $36$.

The solution also shows an equivalent setup with reversed subtraction; area is positive, so the final value remains the same.