question_answer

Let A(\[4,\text{ }-4\]) and B(9, 6) be points on the parabola, \[{{y}^{2}}=4x\]. Let C be chosen on the arc AOB of the parabola, where 0 is the origin, such that the area of \[\Delta \,ACB\] is maximum. Then, the area (in sq. units) of \[\Delta \,ACB\], is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

A) 32                                

B)               \[31\frac{3}{4}\]

C)               \[30\frac{1}{2}\]                                   

D)               \[31\frac{1}{4}\]

Correct Answer: D

Solution :

\[\Delta \,ABC\,\,=\,\,\frac{1}{2}\,\left| \begin{align}   & 4\,\,\,\,\,\,-4\,\,\,\,\,\,\,\,1 \\  & 9\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,1 \\  & {{t}^{2\,\,\,\,\,\,}}\,\,\,\,\,2t\,\,\,\,\,\,\,1 \\ \end{align} \right|\] \[D=60+10t-10{{t}^{2}}\] \[\frac{d\,\Delta }{dt}\,=\,0\,\,\Rightarrow \,\,t=\frac{1}{2}\] \[\frac{{{d}^{2}}\,\Delta }{d{{t}^{2}}}\,=\,\,-20\,\,<\,\,0\] \[\therefore \,\,\,\text{ }max\text{ }at\text{ }t=\frac{1}{2}\] max area \[\Delta =65-\frac{5}{2}\]                               \[=\,\,\,\frac{125}{2}\,\,\,=\,\,31\frac{1}{4}\]