C is a 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 $(4, 2)$ is

y = 15 x - 8

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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 $y = \frac{9 x}{4} - 7$
Slope $M = \frac{d y}{d x} = \frac{d y / d t}{d x / d t}$
$M = \frac{3 t^{2} - 3}{2 t}$
$M = \frac{9}{4}$
Equation of tangent at $(4, 2)$ , $M = \frac{9}{4}$
$(y - 2) = \frac{9}{4} (x - 4) \Rightarrow y = \frac{9}{4} x - 7$

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Similar questions

x = t^{2}

y = t^{3} - 3 t

(4, 2)

y^{2} = 8 x

x y = - 1

y^{2} = 8 x

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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