The equation of tangent at the point $(a t^{2}, a t^{3})$ on the curve $a y^{2} = x^{3}$ is

3 t x - 2 y = a t^{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

t x - 3 y = a t^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 t x + 2 y = a t^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $3 t x - 2 y = a t^{3}$
Let the point is $P (a t^{2}, a t^{3})$
Now given curve is $a y^{2} = x^{3}$
differentiating w.r.t $x$
$2 a y \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{2 a y}$
Thus slope of tangent at P is $= \frac{3 a^{2} t^{4}}{2 a^{2} t^{3}} = \frac{3}{2} t$
Therefore, the equation of tangent at P is, $(y - a t^{3}) = \frac{3}{2} t (x - a t^{2})$
$\Rightarrow 3 t x - 2 y = a t^{3}$
Hence, option 'A' is correct.

Suggest Corrections

Similar questions

(a t^{2}, a t^{3})

a y^{2} = x^{3}

Find the equation of normal to the parabola $y^{2} = 4 a x a t (a t^{2}, 2 a t)$ in terms of t, a.

Find the equations of tangent & normal at parametric point 'P' of the parabola $y^{2} = 4 a x .$

[a t_{1} t_{2}, a (t_{1} + t_{2})], [a t_{2} t_{3}, a (t_{2} + t_{3})], [a t_{3} t_{1}, a (t_{3} + t_{1})]

x = {a t}^{3}, y = {a t}^{4}

Join BYJU'S Learning Program