C is curve defined by the parametric equation $x = t^{2}$ and $y = t^{3} - 3 t$ . The equation of the tangent line to this curve at $(4, 2)$ at a constant speed is

y - 15 x - 58

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y = 2

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y = x - 2

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y = \frac{9 x}{4} - 7

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Solution

The correct option is D $y = \frac{9 x}{4} - 7$
$x = t^{2} y = t^{3} - 3 t y = t (t^{2} - 3) y = \sqrt{x} (x - 3) y = x^{\frac{3}{2}} - 3 x^{\frac{1}{2}}$
For equation of tangent
$\frac{d y}{d x} = \frac{3}{2} x^{\frac{1}{2}} - \frac{3}{2} x^{- \frac{1}{2}} \frac{d y}{d x} = 3 - \frac{1}{4} = \frac{9}{4}$
Equation tangent
$(y - 2) = \frac{9}{4} (x - 4) \Rightarrow y = \frac{9 x}{4} - 7 y = \frac{9 x}{4} - 7$

Suggest Corrections

Similar questions

x = t^{2}

y = t^{3} - 3 t

(4, 2)

y = x e^{x}

List-I	List-II
a) Equation of tangent to the curve $y = b e^{- x / a}$ at $x = 0$	1) $x - 2 y = 2$
b) Equation of tangent to the curve $y = x^{2} + 1$ at $(1, 2)$	2) $y = 2 x$
c) Equation of normal to the curve $y = 2 x - x^{2}$ at $(2, 0)$	3) $x - y = π$
d) Equation of normal to the curve $y = sin x$ at $x = π$	4) $\frac{x}{a} + \frac{y}{b} = 1$

y = x^{2} - 2 x + 7

5 y - 15 x = 13.

y = f (x)

x = (t^{2} + t + 1), y = (t^{2} - t + 1)

t > 0

(1, 1)

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