The equation of the normal to the curve y^4=a | vedclass

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

Question

(C) Given curve is $y^4=a x^3$.On differentiating with respect to $x$,we get:$4 y^3 \frac{d y}{d x} = 3 a x^2$.At the point $(a, a)$,the slope of the

Vedclass · Accepted Answer

$4 x+3 y=7 a$

Vedclass · Answer

(C) Given curve is $y^4=a x^3$.On differentiating with respect to $x$,we get:$4 y^3 \frac{d y}{d x} = 3 a x^2$.At the point $(a, a)$,the slope of the tangent is:$\left(\frac{d y}{d x}ight)_{(a, a)} = \frac{3 a(a)^2}{4(a)^3} = \frac{3 a^3}{4 a^3} = \frac{3}{4}$.The slope of the normal is the negative reciprocal of the tangent slope:$m_{	ext{normal}} = -\frac{1}{3/4} = -\frac{4}{3}$.The equation of the normal at point $(a, a)$ is given by $y - y_1 = m(x - x_1)$:$y - a = -\frac{4}{3}(x - a)$.Multiplying by $3$: $3y - 3a = -4x + 4a$.Rearranging the terms: $4x + 3y = 7a$.

The equation of the normal to the curve $y^4=a x^3$ at $(a, a)$ is

Similar Questions

If the tangent and normal drawn to the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $P\left(\theta=\frac{\pi}{2}\right)$ cut the $X$-axis at $A$ and $B$ respectively,then the area (in sq. units) of $\triangle P A B$ is

The length of the normal at any point to the curve $y=c \cosh \left(\frac{x}{c}\right)$ is

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

The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2xy-3y^2=0$ at the point $(2,2)$ is

The chord of the curve $y=x^{2}+2ax+b$ joining the points where $x=\alpha$ and $x=\beta$ is parallel to the tangent to the curve at abscissa $x$ equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module