Find the equation of the line having slope 0 | vedclass

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

Question

(A) The equation of the given curve is $y=\frac{1}{x^{2}-2x+3}$.The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative

Vedclass · Accepted Answer

$y=\frac{1}{2}$

Vedclass · Answer

(A) The equation of the given curve is $y=\frac{1}{x^{2}-2x+3}$.The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{(x^{2}-2x+3)^{2}} \cdot (2x-2) = \frac{-2(x-1)}{(x^{2}-2x+3)^{2}}$.Given that the slope of the tangent is $0$,we set $\frac{dy}{dx} = 0$:$\frac{-2(x-1)}{(x^{2}-2x+3)^{2}} = 0$.This implies $-2(x-1) = 0$,so $x = 1$.Now,find the corresponding $y$-coordinate by substituting $x=1$ into the curve equation:$y = \frac{1}{(1)^{2}-2(1)+3} = \frac{1}{1-2+3} = \frac{1}{2}$.The point of tangency is $(1, \frac{1}{2})$.The equation of a line with slope $m=0$ passing through $(x_0, y_0)$ is $y - y_0 = m(x - x_0)$.Substituting the values,$y - \frac{1}{2} = 0(x - 1)$,which simplifies to $y = \frac{1}{2}$.

Find the equation of the line having slope $0$ which is tangent to the curve $y=\frac{1}{x^{2}-2x+3}$.

Similar Questions

If the tangent to the curve $y = \frac{x}{x^2-3}$,$x \in R, (x \neq \pm \sqrt{3})$ at a point $(\alpha, \beta) \neq (0,0)$ on it,is parallel to the line $2x + 6y - 11 = 0$,then

The tangent to the curve $y = ax^2 + bx$ at $(2, -8)$ is parallel to the $x$-axis. Then:

The equation of a normal to the curve $x=4 \sec \theta$ and $y=4 \tan^2 \theta$ at $\theta=\frac{\pi}{4}$ is

The ratio between the length of the subtangent at any point other than the origin on the parabola $y^2 = 16ax$ and the abscissa of that point is:

The coordinates of the point$(s)$ on the graph of the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$ where the tangent drawn cuts off intercepts from the coordinate axes which are equal in magnitude but opposite in sign,are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module