Find the equation of the tangent to the curve $y =\sqrt{3x-2}$ which is parallel to the line $4x-2y+5=0$. Also, write the equation of the normal to the curve at the point of contact.

Tangent: $48x−24y−23=0$; Normal: $48x+96y−113=0$

The correct answer is Option (2) → Tangent: $48x−24y−23=0$; Normal: $48x+96y−113=0$

The given curve is $y =\sqrt{3x-2}$   ...(i)

Let P(x1,y1) be a point on (i) the tangent at which is parallel to the line

$4x-2y+5=0$  .....(ii)

Slope of line (ii): $=-\frac{4}{-2}= 2$.

Diff. (i) w.r.t. x, we get

$\frac{dy}{dx}=\frac{1}{2}(3x-2)^{-\frac{1}{2}}.3=\frac{3}{2\sqrt{3x-2}}$,

∴ the slope of tangent to the curve (i) at $P(x_1,y_1) =\frac{3}{2\sqrt{3x_1-2}}$

Since the tangent to the curve (i) at $P(x_1,y_1)$ is parallel to line (ii),

$\frac{3}{2\sqrt{3x_1-2}}=2\sqrt{3x_1-2}=\frac{3}{4}⇒3x_1-2=\frac{9}{16}$

$⇒3x_1=\frac{9}{16}+2⇒3x_1=\frac{41}{16}⇒x_1=\frac{41}{48}$.

As $P(x_1,y_1)$ lies on the curve (i), $y_1 = \sqrt{3x_1-2}⇒y_1=\sqrt{3×\frac{41}{48}-2}=\sqrt{\frac{27}{48}}=\sqrt{\frac{9}{16}}=\frac{3}{4}$

So, the point P is $\left(\frac{41}{48},\frac{3}{4}\right)$.

∴ The equation of tangent to the curve (i) at $P\left(\frac{41}{48},\frac{3}{4}\right)$ is

$y-\frac{3}{4}=2\left(x-\frac{41}{48}\right)$ or $2x-y-\frac{41}{24}+\frac{3}{4}=0$

or $48x−24y−23=0$.

Slope of normal to the curve (i) at the point of contact $P\left(\frac{41}{48},\frac{3}{4}\right)=-\frac{1}{2}$.

$y-\frac{3}{4}=-\frac{1}{2}\left(x-\frac{41}{48}\right)$ or $y-\frac{3}{4}=-\frac{1}{2}x+\frac{41}{96}$

or $96y-72=-48x + 41$ or $48x + 96y - 113 = 0$.