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  • Let be a line which is normal to the curve <img alt="\mathrm{y= 2x^{2}+x+2}" src="https://entrancecor

Let \mathrm{l} be a line which is normal to the curve \mathrm{y= 2x^{2}+x+2} at a point  \mathrm{P} on the curve. If the point \mathrm{Q\left ( 6,4 \right )} lies on the line \mathrm{l} and \mathrm{O} is origin. then the area of the triangle \mathrm{OPQ} is equla to_______.

Option: 1

13


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{y= 2x^{2}+x+2}

\mathrm{Let\; P\, is\left ( h,k \right ).}
\mathrm{\frac{dy}{dx}\mid_{h,k}\; = 4x+1\mid_{h,k}\; = 4h+1= m_{l}}

\mathrm{\Rightarrow m_{N}=\frac{-1}{4 h+1}}

\mathrm{l: y-k=\frac{-1}{4 h+1}(x-h)}

\mathrm{Passes \; through (6,4)=4-k=\frac{-1}{4 h+1}(6-h)}
\mathrm{\Rightarrow(4 h+1)(4-k)=h-6}
\mathrm{\Rightarrow \text { Also } k=2 h^{2}+h+2}
\mathrm{\Rightarrow(4 h+1)\left(4-2 h^{2}-h-2\right)=h-6}
\mathrm{\Rightarrow(4 h+1)\left(2 h^{2}+h-2\right)+h-6=0}
\mathrm{\Rightarrow 8 h^{3}+4 h^{2}-8 h+2 h^{2}+h-2+h-6=0 \Rightarrow 8 h^{3}+6 h^{2}-6 h-8=0}
\mathrm{\Rightarrow 4 h^{3}+3 h^{2}-3 h-4=0 \; \Rightarrow h=1\; \text{satisfy this eqn}}

\mathrm{For \; \; h=1, k=5 \Rightarrow Q(1,5)}

\mathrm{\triangle O P Q=| \frac{1}{2} \begin{vmatrix}1 & 0 & 0 \\ 1 & 6 & 4 \\ 1 & 1 & 5\end{array} |=\frac{1}{2}(30-4)=13}

Posted by

Ritika Harsh

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