The equation of the normal to the curve y(1+x | vedclass

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

Question

(A) Given curve: $y(1+x^{2})=2-x$  $(1)$At the $x$-axis,$y=0$. Substituting $y=0$ in Eq. $(1)$:$0 = 2-x \Rightarrow x=2$.So,the point of intersection

Vedclass · Accepted Answer

$5x-y-10=0$

Vedclass · Answer

(A) Given curve: $y(1+x^{2})=2-x$  $(1)$At the $x$-axis,$y=0$. Substituting $y=0$ in Eq. $(1)$:$0 = 2-x \Rightarrow x=2$.So,the point of intersection is $(2, 0)$.Differentiating Eq. $(1)$ with respect to $x$:$y'(1+x^{2}) + y(2x) = -1$.At point $(2, 0)$,we have:$y'(1+2^{2}) + 0(2 	imes 2) = -1$$y'(5) = -1 \Rightarrow y' = -\frac{1}{5}$.This is the slope of the tangent at $(2, 0)$.The slope of the normal is $-\frac{1}{y'} = -\frac{1}{-1/5} = 5$.The equation of the normal at $(2, 0)$ with slope $m=5$ is:$y - 0 = 5(x - 2)$$y = 5x - 10$$5x - y - 10 = 0$.

The equation of the normal to the curve $y(1+x^{2})=2-x$ at the point where the tangent crosses the $x$-axis is

Similar Questions

The points on the curve $y^2 = x + \sin x$ at which the normal is parallel to the $Y$-axis lie on

The ratio of the length of the subnormal to the square of the length of the subtangent at any point $P$ on the curve $y^2=(2x+1)^3$ is

The slope of the tangent to the curve $y = \int_0^x \frac{1}{1+t^3} dt$ at the point where $x = 1$ is

If the line $y = 4x - 5$ is tangent to the curve $y^2 = ax^3 + b$ at the point $(2, 3)$, then the value of $7a - 2b$ is...

If the tangent at any point on the curve $x^4 + y^4 = a^4$ meets the axes at $p$ and $q$,then the value of $p^{-4/3} + q^{-4/3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module