Question

Find the equation of 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

Let $(h,k)$ be the point on the curve from tangent to be taken $\sqrt{3x-2}$.

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

Now,

$\frac{3}{2\sqrt{3x-2}}=2$ slope of given line (2)

$x=\frac{41}{48}$

Slope of $4x-2y+5=0$, is

$-\frac{coefficientofx}{coeffientofy}$ $=-\frac{4}{-2}$ $=2$

Since, $(h,k)$ on the curve, then point $(h,k)$ satisfies the equation of the curve, thus

$k=\sqrt{3h-2}$

Calculate $k$ when $h=\frac{41}{48}$.

$k=\sqrt{\frac{3\times 41}{48}-2}$

$k=\frac{3}{4}$

Hence, point $(h,k)=(\frac{41}{48},\frac{3}{4})$

Equation of tangent is,

$y-y_1=m(x-x_1)$

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

$48x-24y-23=0$

Equation of normal

$y-y_1=\frac{1}{-m}(x-x_1)$

$y-\frac{3}{4}=\frac{1}{-2}(x-\frac{41}{48})$

$48x+96y=113$