The point on the curve y^{2}=x,the tangent at | vedclass

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

Question

(A) Given curve is $y^{2}=x$. Differentiating both sides with respect to $x$,we get: $2y \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2y}$. The slope

Vedclass · Accepted Answer

$\left(\frac{1}{4}, \frac{1}{2}\right)$

Vedclass · Answer

(A) Given curve is $y^{2}=x$. Differentiating both sides with respect to $x$,we get: $2y \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2y}$. The slope of the tangent is given by $m = 	an(	heta)$. Given $	heta = 45^{\circ}$,so $m = 	an(45^{\circ}) = 1$. Equating the slope,we have: $\frac{1}{2y} = 1 \Rightarrow y = \frac{1}{2}$. Substituting $y = \frac{1}{2}$ into the curve equation $y^{2} = x$: $x = \left(\frac{1}{2}ight)^{2} = \frac{1}{4}$. Thus,the required point is $\left(\frac{1}{4}, \frac{1}{2}ight)$.

The point on the curve $y^{2}=x$,the tangent at which makes an angle $45^{\circ}$ with $X$-axis is

Similar Questions

If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x=12(t+\sin t \cos t)$,$y =12(1+\sin t )^{2}$,$0 < t < \frac{\pi}{2}$,with the positive $x$-axis is $\frac{\pi}{3}$,then $y_{0}$ is equal to

$y=x^2$ is the given curve. Imagine that this curve is dragged along the positive $X$-axis to a distance of '$a$' units. If the acute angle between the curves at two positions is $\theta$, then

At which point on the curve $y = x^2 + 3x$ should the tangent be drawn so that it passes through the point $(0, -9)$?

If the line $ax + by + c = 0$ is a normal to the curve $xy = 1$,then

The length of the normal at any point to the curve $y=c \cosh \left(\frac{x}{c}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module