At which point on the curve y^2 = x does the | vedclass

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

Question

(D) Given the curve is $y^2 = x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,which implies $\frac{dy}{dx} = \frac{1}{2y}$. The

Vedclass · Accepted Answer

$(1/4, 1/2)$

Vedclass · Answer

(D) Given the curve is $y^2 = x$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,which implies $\frac{dy}{dx} = \frac{1}{2y}$. The slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{1}{2y_1}$. Since the tangent makes an angle of $45^{\circ}$ with the $x-$axis,the slope $m = 	an(45^{\circ}) = 1$. Equating the slopes,$\frac{1}{2y_1} = 1$,which gives $y_1 = \frac{1}{2}$. Substituting $y_1 = \frac{1}{2}$ into the curve equation $y^2 = x$,we get $x_1 = (\frac{1}{2})^2 = \frac{1}{4}$. Thus,the required point is $(\frac{1}{4}, \frac{1}{2})$.

At which point on the curve $y^2 = x$ does the tangent make an angle of $45^{\circ}$ with the $x-$axis?

Similar Questions

The length of the normal to the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ at $\theta=\frac{\pi}{2}$ is

Let $\ell$ be a line which is normal to the curve $y=2x^2+x+2$ at a point $P$ on the curve. If the point $Q(6,4)$ lies on the line $\ell$ and $O$ is the origin,then the area of the triangle $OPQ$ is equal to.......

The equation of the tangent to the curve $(\frac{x}{a})^n + (\frac{y}{b})^n = 2$ $(n \in N)$ at the point with abscissa equal to '$a$' is:

The points on the curve $9y^{2} = x^{3}$,where the normal to the curve makes equal intercepts with the axes are

The normal at the point $(1,1)$ on the curve $2y + x^{2} = 3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module