The point on the curve y^{2}=x where the tang | vedclass

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

Question

(B) Given the curve equation: $y^{2} = x$  $(1)$Slope of the tangent $m$ is given by $\frac{dy}{dx}$.We know that the slope $m = 	an(	heta)$,where $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the curve equation: $y^{2} = x$  $(1)$Slope of the tangent $m$ is given by $\frac{dy}{dx}$.We know that the slope $m = 	an(	heta)$,where $	heta$ is the angle with the $x$-axis.Here,$	heta = \frac{\pi}{4}$,so $m = 	an(\frac{\pi}{4}) = 1$.Differentiating equation $(1)$ with respect to $x$:$2y \frac{dy}{dx} = 1$Substituting $m = \frac{dy}{dx} = 1$:$2y(1) = 1 \Rightarrow y = \frac{1}{2}$.Substituting $y = \frac{1}{2}$ into equation $(1)$:$(\frac{1}{2})^{2} = x \Rightarrow x = \frac{1}{4}$.Thus,the required point is $(\frac{1}{4}, \frac{1}{2})$.

The point on the curve $y^{2}=x$ where the tangent makes an angle of $\pi / 4$ with the $x$-axis is

Similar Questions

If the subnormal at any point on the curve $y^n = ax$ is constant,then the value of $n$ is:

The length of the perpendicular drawn from the origin on the normal to the curve $x^2+2xy-3y^2=0$ at the point $(2,2)$ is

If $A = \{P(\alpha, \beta) \mid \text{the tangent drawn at } P \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line}\}$ and $B = \{Q(a, b) \mid \text{the tangent drawn at } Q \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line}\}$, then $n(A) + n(B) = $

The lines tangent to the curves $y^3 - x^2y + 5y - 2x = 0$ and $x^4 - x^3y^2 + 5x + 2y = 0$ at the origin intersect at an angle $\theta$ equal to

For the curve $y=4x^{3}-2x^{5}$,find all the points at which the tangent passes through the origin.

Vedclass Test Series

Exam Paper Generator

Online Exam Module