The area (in sq. units) in the first quadrant bounded by the parabola, $y=x^{2}+1$, the tangent to it at the…

The area (in sq. units) in the first quadrant bounded by the parabola, $y=x^{2}+1$, the tangent to it at the point (2,5) and the coordinate axes is :
  1. $\frac{8}{3}$
  2. $\frac{37}{24}$
  3. $\frac{187}{24}$
  4. $\frac{14}{3}$

Solution


The equation of parabola $x^{2}=y-1$ The equation of tangent at (2,5) to parabola is $y-5=\left(\frac{d y}{d x}\right)_{(2,5)}(x-2)$ $y-5=4(x-2)$ $4 x-y=3$ Then, the required area $=\int_{0}^{2}\left\{\left(x^{2}+1\right)-(4 x-3)\right\} d x-$ Area of $\Delta A O D$ $=\int_{0}^{2}\left(x^{2}-4 x+4\right) d x-\frac{1}{2} \times \frac{3}{4} \times 3$ $=\left[\frac{(x-2)^{3}}{3}\right]_{0}^{2}-\frac{9}{8}=\frac{37}{24}$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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