Find the equation of the normal at the point | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the given curve is $a y^{2}=x^{3}$.Differentiating both sides with respect to $x$,we get:$2 a y \frac{d y}{d x} = 3 x^{2}$$\frac{d

Vedclass · Accepted Answer

$2 x+3 m y-a m^{2}(2+3 m^{2})=0$

Vedclass · Answer

(A) The equation of the given curve is $a y^{2}=x^{3}$.Differentiating both sides with respect to $x$,we get:$2 a y \frac{d y}{d x} = 3 x^{2}$$\frac{d y}{d x} = \frac{3 x^{2}}{2 a y}$.The slope of the tangent at the point $(a m^{2}, a m^{3})$ is:$m_{t} = \left. \frac{d y}{d x} \right|_{(a m^{2}, a m^{3})} = \frac{3(a m^{2})^{2}}{2 a(a m^{3})} = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} = \frac{3 m}{2}$.The slope of the normal at $(a m^{2}, a m^{3})$ is $m_{n} = -\frac{1}{m_{t}} = -\frac{2}{3 m}$.The equation of the normal at $(a m^{2}, a m^{3})$ is given by:$y - a m^{3} = -\frac{2}{3 m}(x - a m^{2})$$3 m y - 3 a m^{4} = -2 x + 2 a m^{2}$$2 x + 3 m y - 3 a m^{4} - 2 a m^{2} = 0$$2 x + 3 m y - a m^{2}(2 + 3 m^{2}) = 0$.

Find the equation of the normal at the point $(a m^{2}, a m^{3})$ for the curve $a y^{2}=x^{3}$.

Similar Questions

For the curve $b{y^2} = {(x + a)^3}$,the square of the subtangent is proportional to:

The coordinates of the point on the curve $9y^2 = x^3$ where the normal to the curve makes equal intercepts with the axes are:

If the tangent drawn at the point $(x_1, y_1)$,where $x_1, y_1 \in \mathbb{N}$,on the curve $y = x^4 - 2x^3 + x^2 + 5x$ passes through the origin,then $x_1 + y_1 =$

Find the length of the tangent and the sub-tangent for the curve $y^2 = 4ax$ at the point $(at^2, 2at)$.

The coordinates of the point$(s)$ on the graph of the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$ where the tangent drawn cuts off intercepts from the coordinate axes which are equal in magnitude but opposite in sign,are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module