The equation of the tangent to the curve y -2 t 2- t - x - t 2 at 1-3 isA- y-x-2 B- 2 y - x -5C- 2 y -3 x -3D- 2 y -5 x -1

The correct option is D 2y = 5x + 1y = 2t^2 + t , x = t^2 Given poin (1, 3) nbsp; t^2 = 1 and 2t^2 + t = 3 ⇒ t = 1 So differentiating w.r.t. t at t = 1 dy/ ...

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Standard XII

Mathematics

The equation ...

Question

The equation of the tangent to the curve $y = 2 t^{2} + t, x = t^{2}$ at $(1, 3)$ is

y = x + 2

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2 y = x + 5

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2 y = 3 x + 3

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2 y = 5 x + 1

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Solution

The correct option is D $2 y = 5 x + 1$
$y = 2 t^{2} + t, x = t^{2}$
Given poin $(1, 3)$
$t^{2} = 1$ and $2 t^{2} + t = 3$
$\Rightarrow t = 1$
So differentiating w.r.t. $t$ at $t = 1$
$\frac{d y}{d t} = 4 t + 1 = 5$
$\frac{d x}{d t} = 2 t = 2$
$∴ \frac{d y}{d x} = \frac{5}{2}$
Equation of tangent,
$\frac{y - 3}{x - 1} = \frac{5}{2} \Rightarrow 2 y - 6 = 5 x - 5 \Rightarrow 2 y = 5 x + 1$

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The equation of the tangent to the curve y=2t2+t , x=t2 at (1,3) is

The correct option is D 2y=5x+1y=2t2+t , x=t2 Given poin (1,3) t2=1 and 2t2+t=3 ⇒t=1 So differentiating w.r.t. t at t=1 dydt=4t+1=5 dxdt=2t=2 ∴dydx=52 Equation of tangent, y−3x−1=52⇒2y−6=5x−5⇒2y=5x+1

The equation of the tangent to the curve $y = 2 t^{2} + t, x = t^{2}$ at $(1, 3)$ is