Observe the following statements for the curve $x = a t^{3}$ , $y = a t^{4}$ at $t = 1$ .
I : The equation of the tangent to the curve is $4 x - 3 y - a = 0$
II : The equation of the normal to the curve is $3 x + 4 y - 7 a = 0$
III: Angle between tangent and normal at any point on the curve is $\frac{π}{2}$
Which of the above statements are correct.

I and II

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II and III

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I and III

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I, II, III

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Solution

The correct option is D I, II, III
$\frac{d y}{d x} = \frac{4 a t^{3}}{3 a t^{2}}$
$\frac{d y}{d x} = \frac{4 t}{3}$
$\frac{d y}{d x} |_{t = 1} = \frac{4}{3}$
$(y - a) = \frac{4}{3} (x - a)$ (tangent)
$4 x - 3 y - a = 0$
$(y - a) = - \frac{3}{4} (x - a)$ (Normal)
$4 y + 3 x - 7 a = 0$
Tangent and normal are always at an angle of $π / 2$ to each other

Suggest Corrections

Similar questions

y = 2. e^{\frac{- x}{3}}

\frac{- 2}{3}

3 x + 2 y + 4 = 0

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 = \sqrt{3 x - 2}

4 x - 2 y + 5 = 0.

x = s i n 3 t

y = c o s 2 t

t = \frac{π}{4}

3 x^{3} - 4 x + 7

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