Let T be the tangent to the ellipse E : x2 + 4y2 = 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 ( 1 \over \sqrt 5 ) , then | \alpha + \beta + \gamma | is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q41 with Solution

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Correct answer: 1.25

Equation of the ellipse and tangent

The ellipse is $x^2+4y^2=5.$

Differentiate implicitly: $2x+8y\frac{dy}{dx}=0$ so $\frac{dy}{dx}=-\frac{x}{4y}.$

At the point $P(1,1)$ , the slope is $m=-\frac{1}{4}.$

Hence the tangent at $(1,1)$ is $y-1=-\frac14(x-1),$ that is, $y=\frac54-\frac{x}{4}.$

Upper branch of the ellipse

From $x^2+4y^2=5,$ we get the upper half as $y=\frac12\sqrt{5-x^2}.$

For $x\in[1,\sqrt5]$ , the tangent lies above the ellipse, so the required area is $A=\int_1^{\sqrt5}\left(\frac54-\frac{x}{4}-\frac12\sqrt{5-x^2}\right)dx.$

Split the integral

$A=\int_1^{\sqrt5}\left(\frac54-\frac{x}{4}\right)dx-\frac12\int_1^{\sqrt5}\sqrt{5-x^2}\,dx.$

First integral

n=\left[\frac54x-\frac{x^2}{8}\right]_1^{\sqrt5}.$$ Evaluating, $$=\left(\frac{5\sqrt5}{4}-\frac{5}{8}\right)-\left(\frac54-\frac18\right) =\frac{5\sqrt5}{4}-\frac{7}{4}.$$ ### Second integral Use the standard result $$\int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right).$$ Here $a=\sqrt5$, so $$\int \sqrt{5-x^2}\,dx=\frac{x}{2}\sqrt{5-x^2}+\frac52\sin^{-1}\left(\frac{x}{\sqrt5}\right).$$ Thus $$\int_1^{\sqrt5}\sqrt{5-x^2}\,dx =\left[\frac{x}{2}\sqrt{5-x^2}+\frac52\sin^{-1}\left(\frac{x}{\sqrt5}\right)\right]_1^{\sqrt5}.$$ At $x=\sqrt5$: $$\frac{x}{2}\sqrt{5-x^2}=0,\qquad \sin^{-1}(1)=\frac\pi2.$$ So value is $$\frac52\cdot\frac\pi2=\frac{5\pi}{4}.$$ At $x=1$: $$\frac12\sqrt{4}=1,$$ and $$\frac52\sin^{-1}\left(\frac1{\sqrt5}\right).$$ Hence $$\int_1^{\sqrt5}\sqrt{5-x^2}\,dx =\frac{5\pi}{4}-1-\frac52\sin^{-1}\left(\frac1{\sqrt5}\right).$$ Therefore $$\frac12\int_1^{\sqrt5}\sqrt{5-x^2}\,dx =\frac{5\pi}{8}-\frac12-\frac54\sin^{-1}\left(\frac1{\sqrt5}\right).$$ 4. **Area** So, $$A=\left(\frac{5\sqrt5}{4}-\frac74\right)-\left(\frac{5\pi}{8}-\frac12-\frac54\sin^{-1}\left(\frac1{\sqrt5}\right)\right).$$ Thus $$A=\frac{5\sqrt5}{4}-\frac54-\frac{5\pi}{8}+\frac54\sin^{-1}\left(\frac1{\sqrt5}\right).$$ Now use $$\sin^{-1}u+\cos^{-1}u=\frac\pi2.$$ So $$\sin^{-1}\left(\frac1{\sqrt5}\right)=\frac\pi2-\cos^{-1}\left(\frac1{\sqrt5}\right).$$ Substitute: $$A=\frac{5\sqrt5}{4}-\frac54-\frac{5\pi}{8}+\frac54\left(\frac\pi2-\cos^{-1}\left(\frac1{\sqrt5}\right)\right).$$ Since $$\frac54\cdot\frac\pi2=\frac{5\pi}{8},$$ these terms cancel, giving $$A=\frac54\sqrt5-\frac54-\frac54\cos^{-1}\left(\frac1{\sqrt5}\right).$$ So comparing with $$\alpha\sqrt5+\beta+\gamma\cos^{-1}\left(\frac1{\sqrt5}\right),$$ we get $$\alpha=\frac54,\qquad \beta=-\frac54,\qquad \gamma=-\frac54.$$ Hence $$|\alpha+\beta+\gamma|=\left|\frac54-\frac54-\frac54\right|=\frac54=1.25.$$ 5. **Comparison with stored answer** Derived answer is $1.25$, which matches the stored correct answer.

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View written solutionFree

Correct answer: 1.25

Equation of the ellipse and tangent

The ellipse is $x^2+4y^2=5.$

Differentiate implicitly: $2x+8y\frac{dy}{dx}=0$ so $\frac{dy}{dx}=-\frac{x}{4y}.$

At the point $P(1,1)$ , the slope is $m=-\frac{1}{4}.$

Hence the tangent at $(1,1)$ is $y-1=-\frac14(x-1),$ that is, $y=\frac54-\frac{x}{4}.$

Upper branch of the ellipse

From $x^2+4y^2=5,$ we get the upper half as $y=\frac12\sqrt{5-x^2}.$

For $x\in[1,\sqrt5]$ , the tangent lies above the ellipse, so the required area is $A=\int_1^{\sqrt5}\left(\frac54-\frac{x}{4}-\frac12\sqrt{5-x^2}\right)dx.$

Split the integral

$A=\int_1^{\sqrt5}\left(\frac54-\frac{x}{4}\right)dx-\frac12\int_1^{\sqrt5}\sqrt{5-x^2}\,dx.$

First integral

n=\left[\frac54x-\frac{x^2}{8}\right]_1^{\sqrt5}.$$ Evaluating, $$=\left(\frac{5\sqrt5}{4}-\frac{5}{8}\right)-\left(\frac54-\frac18\right) =\frac{5\sqrt5}{4}-\frac{7}{4}.$$ ### Second integral Use the standard result $$\int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right).$$ Here $a=\sqrt5$, so $$\int \sqrt{5-x^2}\,dx=\frac{x}{2}\sqrt{5-x^2}+\frac52\sin^{-1}\left(\frac{x}{\sqrt5}\right).$$ Thus $$\int_1^{\sqrt5}\sqrt{5-x^2}\,dx =\left[\frac{x}{2}\sqrt{5-x^2}+\frac52\sin^{-1}\left(\frac{x}{\sqrt5}\right)\right]_1^{\sqrt5}.$$ At $x=\sqrt5$: $$\frac{x}{2}\sqrt{5-x^2}=0,\qquad \sin^{-1}(1)=\frac\pi2.$$ So value is $$\frac52\cdot\frac\pi2=\frac{5\pi}{4}.$$ At $x=1$: $$\frac12\sqrt{4}=1,$$ and $$\frac52\sin^{-1}\left(\frac1{\sqrt5}\right).$$ Hence $$\int_1^{\sqrt5}\sqrt{5-x^2}\,dx =\frac{5\pi}{4}-1-\frac52\sin^{-1}\left(\frac1{\sqrt5}\right).$$ Therefore $$\frac12\int_1^{\sqrt5}\sqrt{5-x^2}\,dx =\frac{5\pi}{8}-\frac12-\frac54\sin^{-1}\left(\frac1{\sqrt5}\right).$$ 4. **Area** So, $$A=\left(\frac{5\sqrt5}{4}-\frac74\right)-\left(\frac{5\pi}{8}-\frac12-\frac54\sin^{-1}\left(\frac1{\sqrt5}\right)\right).$$ Thus $$A=\frac{5\sqrt5}{4}-\frac54-\frac{5\pi}{8}+\frac54\sin^{-1}\left(\frac1{\sqrt5}\right).$$ Now use $$\sin^{-1}u+\cos^{-1}u=\frac\pi2.$$ So $$\sin^{-1}\left(\frac1{\sqrt5}\right)=\frac\pi2-\cos^{-1}\left(\frac1{\sqrt5}\right).$$ Substitute: $$A=\frac{5\sqrt5}{4}-\frac54-\frac{5\pi}{8}+\frac54\left(\frac\pi2-\cos^{-1}\left(\frac1{\sqrt5}\right)\right).$$ Since $$\frac54\cdot\frac\pi2=\frac{5\pi}{8},$$ these terms cancel, giving $$A=\frac54\sqrt5-\frac54-\frac54\cos^{-1}\left(\frac1{\sqrt5}\right).$$ So comparing with $$\alpha\sqrt5+\beta+\gamma\cos^{-1}\left(\frac1{\sqrt5}\right),$$ we get $$\alpha=\frac54,\qquad \beta=-\frac54,\qquad \gamma=-\frac54.$$ Hence $$|\alpha+\beta+\gamma|=\left|\frac54-\frac54-\frac54\right|=\frac54=1.25.$$ 5. **Comparison with stored answer** Derived answer is $1.25$, which matches the stored correct answer.

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