Question

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:

• A. \( \frac{7}{3} \)
• B. 2
• C. \( \frac{8}{3} \)
• D. 3

Hint:

To find the equation of a tangent line, use the point of tangency and solve for the slope using the discriminant.

Answer 1

The Correct Option is C

Tangent
The equation of the tangent is given by: \[ y = m(x + 2) \] Substituting \( y^2 = x - 2 \) into the equation: \[ (m(n + 2))^2 = n - 2 \] This leads to the quadratic equation: \[ m^2x^2 + (4m^2 - 1)x + (4m^2 + 2) = 0 \] Now, for the discriminant to be zero (since it's a tangent line): \[ D = 0 \quad \Rightarrow \quad (4m^2 - 1)^2 - 4m^2(4m^2 + 2) = 0 \] Solving for \( m \): \[ m = \frac{1}{4} \] Now substituting into the equation for \( y \): \[ y = \frac{1}{4}(n + 2) \] The point of tangency is \( (6, 2) \). Now, calculate the area: \[ A = \int_0^2 \left( (y^2 + 2) - (4y - 2) \right) dy \] After solving the integral: \[ A = \frac{8}{3} \]
Thus, the correct option is \( (3) \).

Answer 2

Step 1: The slope of the tangent is given by:

\[ m_T = \frac{1}{2t}. \]

From the equation, we have:

\[ \frac{t}{t^2 + 4} = \frac{1}{2t}. \]

By simplifying this, we get:

\[ 2t^2 = t^2 + 4. \]

Which further simplifies to:

\[ t^2 = 4. \]

Step 2: Now, the area \( A \) is given by:

\[ A = \int_0^2 \left( (y^2 + 2) - (4y - 2) \right) \, dy. \]

On solving this integral, we get:

\[ A = \left[ \frac{(y - 2)^3}{3} \right]_0^2 = \frac{8}{3}. \]