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JEE Main 20204 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2020 · 4 September · Evening Shift · Mathematics · Straight Lines

Let PQ be a diameter of the circle x ^ 2 + y ^ 2 = 9 . If α and β...

Let PQPQ be a diameter of the circle x2+y2=9{x}^{2} + {y}^{2} = 9. If α\alpha and β\beta are the lengths of the perpendiculars from PP and QQ on the straight line, x+y=2x + y = 2 respectively, then the maximum value of αβ\alpha\beta is _______

Correct answer

0

Step-by-step solution

Let P(3cosθ,3sinθ)∴Q(−3cosθ,−3sinθ)P ( 3cos\theta, 3sin\theta ) ∴ Q ( - 3cos\theta, - 3sin\theta )
given line x+y−2=0x + y - 2 = 0 ∴α=∣3cosθ+3sinθ−2∣2∴ \alpha = \dfrac{| 3cos\theta + 3sin\theta - 2 |}{\sqrt{2}}
β=∣−3cosθ−3sinθ−2∣2\beta = \dfrac{| - 3cos\theta - 3sin\theta - 2 |}{\sqrt{2}}
∴αβ=∣(3cosθ+3sinθ−2)⋅(3cosθ+3sinθ+2)2∣=∣9(1+sin⁡2θ)−42∣∴ \alpha\beta = \left|\dfrac{( 3cos\theta + 3sin\theta - 2 ) \cdot ( 3cos\theta + 3sin\theta + 2 )}{2}\right| = \left|\dfrac{9 ( 1 + \sin 2\theta ) - 4}{2}\right|
∴∴ maximum of αβ=7\alpha\beta = 7
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