question_answer

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) \[\frac{14}{3}\]                         

B) \[\frac{56}{3}\]

C) \[\frac{8}{3}\]              

D) \[\frac{32}{3}\]

Correct Answer: C

Solution :

\[y\,=\,{{x}^{2}}\,-\,1\,\to \,(1)\]

Tangent at P (2, 3) to (1) is \[y\,-\,3\,=\,4\,(x-2)\,as\,{{\left( \frac{dy}{dx} \right)}_{\text{p}}}\,=\,4\]

Required area \[=\,\Delta \text{PTN}\,-\,\text{Area}\,\text{of}\,\text{curve}\,\text{A}\text{.P}\] on v-axis \[=\,\frac{1}{2}\,(\text{NT}\times \text{NP})\,-\,\int_{1}^{3}{\sqrt{y+1}dy}\]

\[=\frac{1}{2}(8\times 2)-\frac{2}{3}\left[ {{(y+1)}^{3/2}} \right]_{-1}^{3}\]

\[=\,8\,-\,\frac{16}{3}=\,\frac{8}{5}\]sq. units.