Question

Find the points on the curve y=$\sqrt{x-3}$ where the tangent is perpendicular to the line $6x+3y-5=0.$

Answer 1

Given curve $y=\sqrt{x-3}$

Given line $6x+3y-5=0$

$\Rightarrow 3y=5-6x$

$\Rightarrow y=\frac{5-6x}{3}$

$=-\frac{6}{3}x+\frac{5}{3}$

$m_2=-\frac{6}{3}=-2$

$m_1\times (-2)=-1$

$\Rightarrow m_1=\frac{1}{2}$

$y=\sqrt{x-3}$

$\frac{dy}{dx}=\frac{1}{2\sqrt{x-3}}$

$\Rightarrow \frac{1}{2\sqrt{x-3}}=\frac{1}{2}$

$\Rightarrow \sqrt{x-3}=1$

$\Rightarrow x-3=1$

$\Rightarrow x=4$

$y=\sqrt{4-3}=1$

$\therefore$ point on the curve is $(4,1)$.