Verified Solution • 2025–26 Session moderate

QJEEM21Ma0018: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0018 Solution: Let T be the tangent to the ellipse E : x 2 + 4y 2 = 5 at the…

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 = 5\sqrt 5 is α\alpha 5\sqrt 5 + β\beta + γ\gamma cos −- 1 (15)\left( {{1 \over {\sqrt 5 }}} \right) , then | α\alpha + β\beta + γ\gamma | is equal to ______________.

Step-by-Step Verified Solution

JEE Main 2021 (Online) 20th July Morning Shift Mathematics - Area Under The Curves Question 90 English Explanation
E : x2 + 4y2 = 5

Tangent at P : x + 4y = 5

Required area

=∫15(5−x4−5−x22)dx= \int\limits_1^{\sqrt 5 } {\left( {{{5 - x} \over 4} - {{\sqrt {5 - {x^2}} } \over 2}} \right)dx}

=[5x4−x28−x45−x2−52sin⁡−1x5]15= \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 }

=545−54−54cos⁡−1(15)= {5 \over 4}\sqrt 5 - {5 \over 4} - {5 \over 4}{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)

If we assume α\alpha , β\beta , γ\gamma , ∈\in Q (Not given in question) then α\alpha = 54{5 \over 4} , β\beta = −- 54{5 \over 4} & γ\gamma = −- 54{5 \over 4}

| α\alpha + β\beta + γ\gamma | = 1.25

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring