Let T be the tangent to the ellipse E: x^{2}+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the ellipse is $E: x^{2}+4 y^{2}=5$. The tangent $T$ at $P(1,1)$ is given by $x(1)+4y(1)=5$,which simplifies to $x+4y=5$,or $y = \

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(A) The equation of the ellipse is $E: x^{2}+4 y^{2}=5$. The tangent $T$ at $P(1,1)$ is given by $x(1)+4y(1)=5$,which simplifies to $x+4y=5$,or $y = \frac{5-x}{4}$.The area of the region bounded by the tangent $T$,the ellipse $E$,and the lines $x=1$ and $x=\sqrt{5}$ is given by the integral:$A = \int_{1}^{\sqrt{5}} \left( \frac{5-x}{4} - \frac{\sqrt{5-x^{2}}}{2} \right) dx$Evaluating the integral:$A = \left[ \frac{5x}{4} - \frac{x^{2}}{8} - \frac{1}{2} \left( \frac{x}{2} \sqrt{5-x^{2}} + \frac{5}{2} \sin^{-1} \left( \frac{x}{\sqrt{5}} \right) \right) \right]_{1}^{\sqrt{5}}$$A = \left[ \frac{5x}{4} - \frac{x^{2}}{8} - \frac{x}{4} \sqrt{5-x^{2}} - \frac{5}{4} \sin^{-1} \left( \frac{x}{\sqrt{5}} \right) \right]_{1}^{\sqrt{5}}$At $x=\sqrt{5}$: $\frac{5\sqrt{5}}{4} - \frac{5}{8} - 0 - \frac{5}{4} \sin^{-1}(1) = \frac{10\sqrt{5}-5}{8} - \frac{5\pi}{8}$At $x=1$: $\frac{5}{4} - \frac{1}{8} - \frac{1}{4} \sqrt{4} - \frac{5}{4} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{10-1-4}{8} - \frac{5}{4} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{5}{8} - \frac{5}{4} \sin^{-1} \left( \frac{1}{\sqrt{5}} \right)$Using $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,we have $\sin^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{\pi}{2} - \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$.$A = \left( \frac{10\sqrt{5}-5}{8} - \frac{5\pi}{8} \right) - \left( \frac{5}{8} - \frac{5}{4} \left( \frac{\pi}{2} - \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \right) \right)$$A = \frac{10\sqrt{5}-10}{8} - \frac{5\pi}{8} + \frac{5\pi}{8} - \frac{5}{4} \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) = \frac{5\sqrt{5}}{4} - \frac{5}{4} - \frac{5}{4} \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$Comparing with $\alpha \sqrt{5} + \beta + \gamma \cos^{-1} \left( \frac{1}{\sqrt{5}} \right)$,we get $\alpha = \frac{5}{4}$,$\beta = -\frac{5}{4}$,$\gamma = -\frac{5}{4}$.$|\alpha + \beta + \gamma| = |\frac{5}{4} - \frac{5}{4} - \frac{5}{4}| = |-\frac{5}{4}| = 1.25$.

Let $T$ be the tangent to the ellipse $E: x^{2}+4 y^{2}=5$ at the point $P(1,1)$. If the area of the region bounded by the tangent $T$,ellipse $E$,lines $x=1$ and $x=\sqrt{5}$ is $\alpha \sqrt{5}+\beta+\gamma \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$,then $|\alpha+\beta+\gamma|$ is equal to $....$

Similar Questions

The area (in sq. units) bounded by the curves $y=\sqrt{x}$,$2y-x+3=0$,$X$-axis and lying in the first quadrant,is

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

The area of the region bounded by the parabola $y=x^{2}-4x+5$ and the straight line $y=x+1$ is

Area of the region bounded by the curves $y = \sin x$ and $y = x$ between the lines $x = 0$ and $x = 2\pi$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module