Maths Q19 – JEE-MAIN 2025

A line passing through the point A

(- 2, 0)

, touches the parabola P:

y^{2} = x - 2

at the point B in the first quadrant. The area, of the region bounded by the line AB, parabola P and the x-axis, is:

Clue 1 of 3: Identify the Tangent Point

First, determine the coordinates of the point B where the line AB touches the parabola.

🧠 Full Solution Path

Step 1: Find the point of tangency B and the line equation✦ Active

Let the point of tangency be $B (x_{1}, y_{1})$ . The equation of the parabola is $y^{2} = x - 2$ . The equation of the tangent at $(x_{1}, y_{1})$ is $y y_{1} = \frac{1}{2} (x + x_{1}) - 2$ . Since this line passes through $A (- 2, 0)$ , substitute these coordinates:

0 \cdot y_{1} = \frac{1}{2} (- 2 + x_{1}) - 2

This simplifies to $0 = - 1 + \frac{1}{2} x_{1} - 2 ⟹ \frac{1}{2} x_{1} = 3 ⟹ x_{1} = 6$ . Substitute $x_{1} = 6$ into the parabola equation: $y_{1}^{2} = 6 - 2 = 4 ⟹ y_{1} = \pm 2$ . Since B is in the first quadrant, $y_{1} = 2$ . So, $B = (6, 2)$ . The equation of the line AB passing through $A (- 2, 0)$ and $B (6, 2)$ is $y - 0 = \frac{2 - 0}{6 - (- 2)} (x - (- 2)) ⟹ y = \frac{1}{4} (x + 2)$ , which can be rewritten as $x = 4 y - 2$ .

Step 2: Set up the integral for the area○ Expand

The region is bounded by the line AB ( $x = 4 y - 2$ ), the parabola P ( $x = y^{2} + 2$ ), and the x-axis ( $y = 0$ ). It is convenient to integrate with respect to $y$ . The limits for $y$ are from $0$ (x-axis) to $2$ (y-coordinate of B). For any given $y$ in this range, the parabola $x_{P} = y^{2} + 2$ is to the right of the line $x_{L} = 4 y - 2$ . The area is given by the integral:

A r e a = \int_{0}^{2} (x_{P} - x_{L}) d y = \int_{0}^{2} ((y^{2} + 2) - (4 y - 2)) d y

Simplifying the integrand, we get:

A r e a = \int_{0}^{2} (y^{2} - 4 y + 4) d y

💡 Teacher's Secret Hint

Remember to integrate with respect to the variable that simplifies the boundaries of the region.

Step 3: Evaluate the integral○ Expand

The integrand is a perfect square, $(y - 2)^{2}$ . Evaluate the definite integral:

A r e a = \int_{0}^{2} (y - 2)^{2} d y = {[\frac{(y - 2)^{3}}{3}]}_{0}^{2}

Substitute the limits of integration:

A r e a = \frac{(2 - 2)^{3}}{3} - \frac{(0 - 2)^{3}}{3} = \frac{0^{3}}{3} - \frac{(- 2)^{3}}{3} = 0 - \frac{- 8}{3} = \frac{8}{3}

💡 Teacher's Secret Hint

Be careful with the signs when evaluating the definite integral.

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Maths Question 19 – JEE-MAIN 2025

Clue 1 of 3: Identify the Tangent Point

🧠 Full Solution Path