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

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

Question

(A) Given the curve equation: $(1 + x^2)y = 2 - x$ ......$(i)$It crosses the $x$-axis where $y = 0$. Substituting $y = 0$ in $(i)$:$0 = 2 - x \Rightar

Vedclass · Accepted Answer

$x + 5y = 2$

Vedclass · Answer

(A) Given the curve equation: $(1 + x^2)y = 2 - x$ ......$(i)$It crosses the $x$-axis where $y = 0$. Substituting $y = 0$ in $(i)$:$0 = 2 - x \Rightarrow x = 2$.So,the curve meets the $x$-axis at the point $(2, 0)$.Now,express $y$ in terms of $x$:$y = \frac{2 - x}{1 + x^2}$.Differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{(1 + x^2)(-1) - (2 - x)(2x)}{(1 + x^2)^2}$$= \frac{-1 - x^2 - 4x + 2x^2}{(1 + x^2)^2} = \frac{x^2 - 4x - 1}{(1 + x^2)^2}$.The slope of the tangent at $(2, 0)$ is:$m = \left. \frac{dy}{dx} \right|_{(2, 0)} = \frac{2^2 - 4(2) - 1}{(1 + 2^2)^2} = \frac{4 - 8 - 1}{25} = \frac{-5}{25} = -\frac{1}{5}$.The equation of the tangent at $(2, 0)$ with slope $m = -\frac{1}{5}$ is:$y - 0 = -\frac{1}{5}(x - 2)$$5y = -x + 2$$x + 5y = 2$.

The equation of the tangent to the curve $(1 + x^2)y = 2 - x$ at the point 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 point$(s)$ on the curve $y^3 + 3x^2 = 12y$ where the tangent is vertical (parallel to $y$-axis) is (are):

Find the condition for the curves $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $xy=c^{2}$ to intersect orthogonally.

If $x^2 + y^2 = R^2$ $(R > 0)$,then $k = \frac{y''}{\sqrt{(1 + (y')^2)^3}}$. The value of $k$ in terms of $R$ alone is equal to:

Find the equation of all lines having slope $2$ which are tangents to the curve $y = \frac{1}{x-3}, x \neq 3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module