JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 20)
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$$\left( {{1 \over {\sqrt 5 }}} \right)$$, then |$$\alpha$$ + $$\beta$$ + $$\gamma$$| is equal to ______________.
Answer
1.25
Explanation
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E : x2 + 4y2 = 5
Tangent at P : x + 4y = 5
Required area
$$ = \int\limits_1^{\sqrt 5 } {\left( {{{5 - x} \over 4} - {{\sqrt {5 - {x^2}} } \over 2}} \right)dx} $$
$$ = \left[ {{{5x} \over 4} - {{{x^2}} \over 8} - {x \over 4}\sqrt {5 - {x^2}} - {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right]_1^{\sqrt 5 }$$
$$ = {5 \over 4}\sqrt 5 - {5 \over 4} - {5 \over 4}{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$
If we assume $$\alpha$$, $$\beta$$, $$\gamma$$, $$\in$$ Q (Not given in question) then $$\alpha$$ = $${5 \over 4}$$, $$\beta$$ = $$-$$$${5 \over 4}$$ & $$\gamma$$ = $$-$$$${5 \over 4}$$
|$$\alpha$$ + $$\beta$$ + $$\gamma$$| = 1.25
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