JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 12)
Let PQ be a diameter of the circle x2 + y2 = 9. If $$\alpha $$ and $$\beta $$ are the lengths of the perpendiculars from P and Q on the straight line,
x + y = 2 respectively, then the maximum value of $$\alpha\beta $$ is _____.
x + y = 2 respectively, then the maximum value of $$\alpha\beta $$ is _____.
Answer
7
Explanation
Let $$P(3\cos \theta ,\,3\sin \theta )$$
$$Q( - 3\cos \theta ,\, - 3\sin \theta )$$
$$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$
$$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$
$$\alpha \beta = \left| {{{{{\left( {3\cos \theta + 3\sin \theta } \right)}^2} - 4} \over 2}} \right|$$
$$ = \left| {{{5 + 9\sin 2\theta } \over 2}} \right|$$
$$\alpha {\beta _{\max }}$$$$ = {{5 + 9} \over 2} = 7$$ (when sin2$$\theta $$ = 1)
$$Q( - 3\cos \theta ,\, - 3\sin \theta )$$
$$\alpha = \left| {{{3\cos \theta + 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$
$$\beta = \left| {{{ - 3\cos \theta - 3\sin \theta - 2} \over {\sqrt 2 }}} \right|$$
$$\alpha \beta = \left| {{{{{\left( {3\cos \theta + 3\sin \theta } \right)}^2} - 4} \over 2}} \right|$$
$$ = \left| {{{5 + 9\sin 2\theta } \over 2}} \right|$$
$$\alpha {\beta _{\max }}$$$$ = {{5 + 9} \over 2} = 7$$ (when sin2$$\theta $$ = 1)
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