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 curves $y=x^2-1$ and $y=8x-x^2-9$:

The tangent to the curve $y = e^{2x}$ at the point $(0, 1)$ meets the $x$-axis at

If $y=4x-5$ is tangent to the curve $y^2=px^3+q$ at $(2,3)$,then

Find all points on the curve $y = 4x^3 - 2x^5$ at which the tangent passes through the origin.

$V$ is the set of points on the curve $y^3 - 3xy + 2 = 0$ where the tangent is vertical. Then $V = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module