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, $x_{1}=2$
$$ \begin{aligned} & \text { and } \quad y_{1}=3 \text {. } \ & \Rightarrow \quad \frac{y+3}{2}=2 x-1 \ & \Rightarrow \quad y=4 x-5 \end{aligned} $$
Now, required area $=$ area of shaded region
$=\int_{0}^{2}(y$ (parabola) $-y($ tangent $)) d x$
$=\int_{0}^{2}\left[\left(x^{2}-1\right)-(4 x-5)\right] d x$
$=\int_{0}^{2}\left(x^{2}-4 x+4\right) d x=\int_{0}^{2}(x-2)^{2} d x$
$=\left|\frac{(x-2)^{3}}{3}\right|_{0}^{2}=\frac{(2-2)^{3}}{3}-\frac{(0-2)^{3}}{3}=\frac{8}{3}$ sq units.
