JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

a > 0,\;b > 0

the maximum area of the triangle formed by the points

O\left( {0,\;0} \right),\;A\left( {a\cos \theta ,\;b\sin \theta } \right)

and

B\left( {a\cos \theta ,\; - b\sin \theta } \right)

is (in sq unit)

View answer and explanation

Correct option: A

\frac{{ab}}{2}\,\mathrm{when}\;\mathrm{\theta } = \frac{\pi }{4}

Option A (

\frac{{ab}}{2}\,\mathrm{when}\;\mathrm{\theta } = \frac{\pi }{4}

) is correct because the maximum area is achieved when

\sin 2\theta = 1

, which occurs at

\theta = \frac{\pi}{4}

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

x-3 y=p, a x+2 y=q

a x+y=r

\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

y = 2\;x + 4

6x = 3y + 5

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

(3,8)

x+3 y=7

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0