Let T be the tangent to the ellipse E: x ^ 2 + 4 y ^ 2 = 5 at the...

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}

\alpha\sqrt{5} + \beta + \gamma\cos^{- 1}\left(\dfrac{1}{\sqrt{5}}\right)

, then

| \alpha + \beta + \gamma |

is equal to______. 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

Correct answer

Step-by-step solution

We have,

${x}^{2} + 4{y}^{2} = 5$

$⇒ \dfrac{{x}^{2}}{5} + \dfrac{{y}^{2}}{\left(\dfrac{5}{4}\right)}$ = 1 $<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b72fcaf8-7c43-470d-acba-aff3bc540cab-image.png" style="width: 388px; height: 200px;" />Tangent at$ P\left(1, 1\right) $to the ellipse is$ x + 4y = 5 $ Required Area$ = {∫}_{1}^{\sqrt{5}}\left(\dfrac{5 - x}{4} - \dfrac{\sqrt{5 - {x}^{2}}}{2}\right)dx $$ = {\left[\dfrac{5x}{4} - \dfrac{{x}^{2}}{8} - \dfrac{x}{4}\sqrt{5 - {x}^{2}} - \dfrac{5}{4}\sin^{- 1}\dfrac{x}{\sqrt{5}}\right]}_{1}^{\sqrt{5}} $$ = \left[\dfrac{5\sqrt{5}}{4} - \dfrac{5}{8} - \dfrac{5}{4} \cdot \left(\dfrac{\pi}{2}\right) - \dfrac{5}{4} + \dfrac{1}{8} + \dfrac{1}{2} + \dfrac{5}{4}\sin^{- 1}\left(\dfrac{1}{\sqrt{5}}\right)\right] $$ = \left[\dfrac{5\sqrt{5}}{4} - \dfrac{5}{4} - \dfrac{5}{4} \cdot \left(\dfrac{\pi}{2}\right) + \dfrac{5}{4}\left(\dfrac{\pi}{2} - \cos^{- 1}\left(\dfrac{1}{\sqrt{5}}\right)\right)\right] $$ = \left[\dfrac{5\sqrt{5}}{4} - \dfrac{5}{4} - \dfrac{5\pi}{8} + \dfrac{5\pi}{8} - \dfrac{5}{4}\cos^{- 1}\left(\dfrac{1}{\sqrt{5}}\right)\right] $$ = \dfrac{5}{4}\sqrt{5} - \dfrac{5}{4} - \dfrac{5}{4}\cos^{- 1}\left(\dfrac{1}{\sqrt{5}}\right) $So,$ \alpha = \dfrac{5}{4}, \beta = - \dfrac{5}{4} \text{ and } \gamma = - \dfrac{5}{4} $\left|\alpha + \beta + \gamma\right| = 1 . 25$

Practice Area Under Curves on Quantrex Academy →

More from Area Under Curves