Let ell be a line which is normal to the curv | vedclass

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(A) The slope of the tangent to the curve $y=2x^2+x+2$ at any point $(h, k)$ is given by $\frac{dy}{dx} = 4x+1$. At point $P(h, k)$,the slope of the t

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$13$

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(A) The slope of the tangent to the curve $y=2x^2+x+2$ at any point $(h, k)$ is given by $\frac{dy}{dx} = 4x+1$. At point $P(h, k)$,the slope of the tangent is $m_t = 4h+1$. The slope of the normal line $\ell$ at $P$ is $m_n = -\frac{1}{4h+1}$. The equation of the normal line $\ell$ passing through $P(h, k)$ is $y-k = -\frac{1}{4h+1}(x-h)$. Since $Q(6, 4)$ lies on $\ell$,we have $4-k = -\frac{1}{4h+1}(6-h)$. Substituting $k = 2h^2+h+2$,we get $4-(2h^2+h+2) = -\frac{6-h}{4h+1}$. $(2-h-2h^2)(4h+1) = h-6$. $8h+2-4h^2-h-8h^3-2h^2 = h-6$. $-8h^3-6h^2+7h+2 = h-6$. $8h^3+6h^2-6h-8 = 0$. Dividing by $2$,$4h^3+3h^2-3h-4 = 0$. $(h-1)(4h^2+7h+4) = 0$. Since $4h^2+7h+4$ has no real roots (discriminant $D = 49-64 Then $k = 2(1)^2+1+2 = 5$. So $P$ is $(1, 5)$. The area of $	riangle OPQ$ with vertices $O(0, 0)$,$P(1, 5)$,and $Q(6, 4)$ is given by: Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. Area $= \frac{1}{2} |0(5-4) + 1(4-0) + 6(0-5)| = \frac{1}{2} |4 - 30| = \frac{1}{2} |-26| = 13$.

Let $\ell$ be a line which is normal to the curve $y=2x^2+x+2$ at a point $P$ on the curve. If the point $Q(6,4)$ lies on the line $\ell$ and $O$ is the origin,then the area of the triangle $OPQ$ is equal to.......

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