SATHEE: Area Question 13

Area Question 13

Question 13

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) $\frac{8}{3}$ (b) $\frac{56}{3}$ (c) $\frac{32}{3}$ (d) $\frac{14}{3}$

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Solution:

Given, equation of parabola is $y = x^{2} - 1$ , which can be rewritten as $x^{2} = y + 1$ or $x^{2} = (y - (- 1))$ .

$\Rightarrow$ Vertex of parabola is $(0, - 1)$ and it is open upward.

Equation of tangent at $(2, 3)$ is given by $T = 0$

$\Rightarrow \frac{y + y_{1}}{2} = x x_{1} - 1$ , where, $x_{1} = 2$

$\begin{aligned} and y_{1} = 3 . & \Rightarrow \frac{y + 3}{2} = 2 x - 1 & \Rightarrow y = 4 x - 5 \end{aligned}$

Now, required area $=$ area of shaded region

$= \int_{0}^{2} (y$ (parabola) $- y ($ tangent $)) d x$

$= \int_{0}^{2} [(x^{2} - 1) - (4 x - 5)] d x$

$= \int_{0}^{2} (x^{2} - 4 x + 4) d x = \int_{0}^{2} (x - 2)^{2} d x$

$= {| \frac{(x - 2)^{3}}{3} |}_{0}^{2} = \frac{(2 - 2)^{3}}{3} - \frac{(0 - 2)^{3}}{3} = \frac{8}{3}$ sq units.

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