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The area (in sq. units) bounded by the p...

The area (in sq. units) bounded by the parabola `y=x^2-1`, the tangent at the point (2,3) to it and the y-axis is

A

`14/3`

B

`56/3`

C

`8/3`

D

`32/3`

Text Solution

AI Generated Solution

The correct Answer is:
To find the area bounded by the parabola \( y = x^2 - 1 \), the tangent line at the point \( (2, 3) \), and the y-axis, we will follow these steps: ### Step 1: Find the slope of the tangent line The equation of the parabola is given as \( y = x^2 - 1 \). To find the slope of the tangent line at the point \( (2, 3) \), we first differentiate the equation of the parabola. \[ \frac{dy}{dx} = 2x \] Now, substituting \( x = 2 \): \[ \text{slope} = 2(2) = 4 \] ### Step 2: Write the equation of the tangent line Using the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency \( (2, 3) \): \[ y - 3 = 4(x - 2) \] Simplifying this: \[ y - 3 = 4x - 8 \implies y = 4x - 5 \] ### Step 3: Find the intersection of the tangent line with the y-axis To find the intersection with the y-axis, set \( x = 0 \): \[ y = 4(0) - 5 = -5 \] Thus, the point of intersection is \( (0, -5) \). ### Step 4: Determine the area between the parabola, tangent line, and y-axis We need to find the area between the parabola \( y = x^2 - 1 \) and the tangent line \( y = 4x - 5 \) from \( x = 0 \) to \( x = 2 \). The area \( A \) can be calculated using the integral: \[ A = \int_{0}^{2} \left( (4x - 5) - (x^2 - 1) \right) dx \] Simplifying the integrand: \[ A = \int_{0}^{2} (4x - 5 - x^2 + 1) dx = \int_{0}^{2} (-x^2 + 4x - 4) dx \] ### Step 5: Evaluate the integral Now we evaluate the integral: \[ A = \int_{0}^{2} (-x^2 + 4x - 4) dx \] Calculating the integral: \[ = \left[ -\frac{x^3}{3} + 2x^2 - 4x \right]_{0}^{2} \] Substituting the limits: \[ = \left( -\frac{(2)^3}{3} + 2(2)^2 - 4(2) \right) - \left( -\frac{(0)^3}{3} + 2(0)^2 - 4(0) \right) \] Calculating the upper limit: \[ = \left( -\frac{8}{3} + 8 - 8 \right) = -\frac{8}{3} \] The lower limit is zero, so: \[ A = -\frac{8}{3} - 0 = -\frac{8}{3} \] Since area cannot be negative, we take the absolute value: \[ A = \frac{8}{3} \] ### Final Answer Thus, the area bounded by the parabola, the tangent line, and the y-axis is: \[ \boxed{\frac{8}{3}} \]
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