Find the point at which the tangent to the cu | vedclass

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

Question

(B) The slope of the tangent to the curve $y = \sqrt{4x - 3} - 1$ is given by the derivative $\frac{dy}{dx}$.$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{4x -

Vedclass · Accepted Answer

$(3, 2)$

Vedclass · Answer

(B) The slope of the tangent to the curve $y = \sqrt{4x - 3} - 1$ is given by the derivative $\frac{dy}{dx}$.$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{4x - 3} - 1) = \frac{1}{2\sqrt{4x - 3}} 	imes 4 = \frac{2}{\sqrt{4x - 3}}$.Given that the slope is $\frac{2}{3}$,we set the derivative equal to $\frac{2}{3}$:$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3}$.$\sqrt{4x - 3} = 3$.Squaring both sides,we get $4x - 3 = 9$.$4x = 12$,which implies $x = 3$.Now,substitute $x = 3$ into the original equation to find $y$:$y = \sqrt{4(3) - 3} - 1 = \sqrt{12 - 3} - 1 = \sqrt{9} - 1 = 3 - 1 = 2$.Thus,the required point is $(3, 2)$.

Find the point at which the tangent to the curve $y = \sqrt{4x - 3} - 1$ has its slope $\frac{2}{3}$.

Similar Questions

The angle between the curves $x^2=8y$ and $xy=8$ is

Find the point on the curve $x = a(\theta + \sin \theta)$,$y = a(1 - \cos \theta)$ where the tangent makes an angle of $\pi / 4$ with the $x$-axis.

The angle between the curves $y=\sin x$ and $y=\cos x$ is

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $(-2,0)$ and cuts the $Y$-axis at a point $Q$ where its gradient is $3$. Then the values of $a, b, c$ respectively are:

The number of those tangents to the curve $y^2 - 2x^3 - 4y + 8 = 0$ which pass through the point $(1, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module