Question

The point of contact of the tangent to the curve $y=\sqrt{x-1}$ , having a slope of $\frac{1}{3}$ is

• A. $(\frac{13}{4},\frac{3}{2})$
• B. $(\frac{29}{4},\frac{5}{2})$
• C. $(\frac{53}{4},\frac{7}{2})$
• D. $(2,1)$

Answer 1

The correct option is A $(\frac{13}{4},\frac{3}{2})$

Given curve $y=\sqrt{x-1}$
Slope of the tangent,
$m=y^{\prime }\Rightarrow m=\frac{1}{2\sqrt{x-1}}\Rightarrow \frac{1}{3}=\frac{1}{2\sqrt{x-1}}\Rightarrow \frac{1}{9}=\frac{1}{4(x-1)}\Rightarrow (x-1)=\frac{9}{4}\Rightarrow x=\frac{13}{4}\therefore y=\sqrt{\frac{13}{4}-1}=\frac{3}{2}$