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

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

Question

(A) Given the curve equation: $(1+x^2)y = 2-x$. To find the point where the curve crosses the $X$-axis,set $y=0$: $(1+x^2)(0) = 2-x \implies 2-x = 0 \

Vedclass · Accepted Answer

$x+5y=2$

Vedclass · Answer

(A) Given the curve equation: $(1+x^2)y = 2-x$. To find the point where the curve crosses the $X$-axis,set $y=0$: $(1+x^2)(0) = 2-x \implies 2-x = 0 \implies x=2$. So,the point of contact is $(2, 0)$. Now,differentiate the equation with respect to $x$: $(1+x^2) \frac{dy}{dx} + y(2x) = -1$. Substitute $x=2$ and $y=0$ to find the slope $m$: $(1+2^2) \frac{dy}{dx} + 0(2 	imes 2) = -1 (5) \frac{dy}{dx} = -1 \frac{dy}{dx} = -\frac{1}{5}$. The equation of the tangent at $(2, 0)$ with slope $m = -\frac{1}{5}$ is: $y - 0 = -\frac{1}{5}(x - 2) 5y = -x + 2 x + 5y = 2$. Thus,the correct option is $A$.

The equation of the tangent to the curve $(1+x^2)y = 2-x$,where it crosses the $X$-axis,is

Similar Questions

The length of the subtangent at any point $(x_1, y_1)$ on the curve $y=5^x$ is

The acute angle between the curves $y=3x^2-2x-1$ and $y=x^3-1$ at their point of intersection which lies in the first quadrant is

The equation of the tangent to the curve $y^{2}=ax^{2}+b$ at the point $(2,3)$ is $y=4x-5$. Then the values of $a$ and $b$ are:

Let $a, b, c \in \mathbb{R}$ be such that $2a + 3b + 6c = 0$ and $g(x)$ be the antiderivative of $f(x) = ax^2 + bx + c$. If the slopes of the tangents drawn to the curve $y = g(x)$ at $(1, g(1))$ and $(2, g(2))$ are equal, then

The angle of intersection of the curves $r = \sin \theta + \cos \theta$ and $r = 2 \sin \theta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module