Area Ques 21
Question
- 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) $\frac{14}{3}$
Show Answer
Answer:
Correct Answer: 21.(a)
Solution:
Formula:
- 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.
BETA
