Question

Find the equation of the tangent to the curve $y=\sqrt{3x-2}$ which is parallel to the line $4x-2y+5=0$.
Further find the equation of the normal to the curve.

Answer 1

Equation of tangent is $\frac{dy}{dx}=\frac{1}{2\sqrt{3x-2}}\times 3at(h,k)\text{slope of tangent}=\frac{3}{2\sqrt{3h-2}}$ Given tangent is parallel to line $4x-2y+5\text{so slope of tangent}=\text{slope of line}$

$\Rightarrow \frac{3}{2\sqrt{3h-2}}=2$

$\Rightarrow 9=16(3h-2)$

$\Rightarrow 3h=2+\frac{9}{16}$

$\Rightarrow h=\frac{41}{48}$

$\because \left(h_{1}k\right)$ lies on curve so

$k=\sqrt{3h-2}=\sqrt{3\times \frac{41}{48}-2}=\sqrt{\frac{9}{16}}=\frac{3}{4}$

$\therefore \left(h_{1}k\right)=(\frac{41}{48},\frac{3}{4})$

$\text{eqn of tangent is}y-\frac{3}{4}=2(x-\frac{41}{48})$

$\Rightarrow 4y-3=\frac{8}{48}(48x-41)$

$\Rightarrow 24y-18=40x-41$

$\Rightarrow 48x-24y=23$

$\text{eqn of normal is}y-\frac{3}{4}=\frac{-1}{2}(x-\frac{41}{48})$

$\Rightarrow 4y-3=\frac{-2}{48}(48x-41)$

$\Rightarrow 96y-72=-48x+41$

$\Rightarrow 48x+96y=113$