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

Correct answer: 0 The given curve is y = 2 x ^ 2 + x + 2 (dy)/(dx) = 4x + 1 Let P be (h, k) , then normal at P is y - k = - (1)/(4h + 1)(x - h) Since this normal passes through Q(6, 4) ∴ 4 - k = - (1)/(4h + 1)(6 - h) ⇒ (4h + 1) (4 - k) + 6 - h = 0 Also k = 2 h ^ 2 + h + 2 ∴ (4h + 1)(4 - 2 h ^ 2 - h - 2) + 6 + h = 0 ⇒ 4 h ^ 3 - 3 h ^ 2 + 3h - 8 = 0 ⇒ h = 1, k = 5 So P ≡ (1, 5) Now area of ∆ OPQ will be = (1)/(2)| cases 1 & 0 & 0 1 & 1 & 5 1 & 6 & 4 cases | = 13

JEE Main 202228 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2022 · 28 June · Morning Shift · Mathematics · Application of Derivatives

Let l be a line which is normal to the curve y = 2 x ^ 2 + x + 2 at...

Let

l be a line which is normal to the curve

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

at a point

on the curve. If the point

Q\left(6, 4\right)

lies on the line

l

and

O

is origin, then the area of the triangle

OPQ

is equal to _____. Let l be a line which is normal to the curve y = 2 x ^ 2 + x + 2 at a point P on the curve. If the point Q(6, 4) lies

Correct answer

Step-by-step solution

The given curve is $y = 2{x}^{2} + x + 2$

$\dfrac{dy}{dx} = 4x + 1$

Let l be a line which is normal to the curve y = 2 x ^ 2 + x + 2 at a point P on the curve. If the point Q(6, 4) lies

Let $P$ be \left(h, k\right) $, then normal at$ P $is$ y - k = - \dfrac{1}{4h + 1}\left(x - h\right)

Since this normal passes through $Q\left(6, 4\right)$$∴ 4 - k = - \dfrac{1}{4h + 1}\left(6 - h\right)$⇒ \left(4h + 1\right)$\left(4 - k\right) + 6 - h = 0$Also $k = 2{h}^{2} + h + 2$$∴ \left(4h + 1\right)\left(4 - 2{h}^{2} - h - 2\right)$ + 6 + h = 0$$⇒ 4{h}^{3} - 3{h}^{2} + 3h - 8 = 0$$⇒ h = 1, k = 5$So $P ≡ \left(1, 5\right)$Now area of $∆ OPQ$ will be $= \dfrac{1}{2}\left|$\begin{cases}1 & 0 & 0 \\ 1 & 1 & 5 \\ 1 & 6 & 4\end{cases}$ \right| $= 13$

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