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

At which points on the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$,is the tangent line: $(i)$ parallel to the $x$-axis? $(ii)$ parallel to the $y$-axis? $(iii)$ making equal angles with both axes?

Let the normal at a point $P$ on the curve $y^{2}-3x^{2}+y+10=0$ intersect the $y$-axis at $(0, \frac{3}{2})$. If $m$ is the slope of the tangent at $P$ to the curve,then $|m|$ is equal to

The equation of the normal to the curve $y = \sin x$ at the point $(0, 0)$ is

Which of the following points lies on the tangent to the curve $x^{4} e^{y}+2 \sqrt{y+1}=3$ at the point $(1,0)$?

The length of the segment of the tangent line to the curve $x=a \cos ^3 t, y=a \sin ^3 t$, at any point on the curve cut off by the coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module