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  • Let <img alt="\mathrm{\mathrm{A}\left(\frac{3}{\sqrt{a}}, \sqrt{a}\right), a>0,}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BA%7D%5Cleft%28%5Cfrac%7B3%7D%7B%5Cs

Let \mathrm{\mathrm{A}\left(\frac{3}{\sqrt{a}}, \sqrt{a}\right), a>0,} be a fixed point in the \mathrm{xy-plane}. The image of \mathrm{A} in \mathrm{y-axis} be \mathrm{B} and the image of \mathrm{B} in \mathrm{x-axis} be C. If  \mathrm{D(3 \cos \theta, a \sin \theta)} is a point in the fourth quadrant such that the maximum area of \mathrm{\triangle \, \mathrm{ACD}}  is 12 square units, then a is equal to _________.

Option: 1

8


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{A\left ( \frac{3}{\sqrt{a}},\sqrt{a} \right )}
\mathrm{B\left ( \frac{-3}{\sqrt{a}},\sqrt{a} \right )}
\mathrm{C\left ( \frac{-3}{\sqrt{a}},-\sqrt{a} \right )}                      \mathrm{D:\left ( 3\cos \theta ,a\sin \theta \right )}

\mathrm{Ar\left ( \triangle \, ACD \right )= \frac{1}{2}\left [ \frac{3}{\sqrt{a}} \left ( -\sqrt{a} -a\sin \theta \right )+\left ( \frac{-3}{\sqrt{a}} \right )\left ( a\sin \theta -\sqrt{a} \right )+3\cos \theta \left ( \sqrt{a}+\sqrt{a} \right )\right ]}
                           \mathrm{= 3\sqrt{a}\left ( \cos \theta -\sin \theta \right )}

Maximum area \mathrm{= 12= 3\sqrt{a}\; \sqrt{2}}
                               \mathrm{ 2\sqrt{a}= \sqrt{a}}
                                  \mathrm{ a= 8}

Posted by

Ritika Kankaria

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