Question

Find the equation of tangent to the curve y=$\sqrt{3x-2}$ which is parallel to the line $4x-2y+=0$. further find the equation of the normal to the curve.

Answer 1

Consider the given equation of the curve,

$y=\sqrt{3x-2}$ ……(1)

differentiate with respect to x ,we get

$\frac{dy}{dx}=-\frac{3}{2\sqrt{3x-2}}$

But Given the equation,

$4x-2y=0$ …..(2)

Slope of equation (2) is

$=-\frac{(-4)}{2}=2$

Given tangent of equation (1) is parallel to equation (2), so

$\begin{align}

$\frac{-3}{2\sqrt{3x-2}}=2$

$\sqrt{3x-2}=\frac{-3}{4}$

$3x-2={\left(\frac{3}{4}\right)}^2$

$3x=\frac{9}{16}$

$x=\frac{3}{16}$

Value of x put in equation (2), we get.

$4\times \frac{3}{16}=2y$

$y=\frac{3}{8}$

Now ,

$y-\frac{3}{8}=2(x-\frac{3}{16})$

$16x-16y=-3$

This is required answer