Question

The slope of tangent(s) to curve $y=\sqrt{18-x^2}$ at the point(s) where its ordinate and abscissa are equal is/are

• A. $-1$
• B. $0$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $-1$

Given $y=x$
$\Rightarrow x=\sqrt{18-x^2}$
$\Rightarrow 2x^2=18$
$\Rightarrow x=3,-3(\text{rejected,}\because x>0)\frac{dy}{dx}=\frac{-2x}{2\sqrt{18-x^2}}=\frac{-x}{\sqrt{18-x^2}}{\frac{dy}{dx}|}_{x=3}=\frac{-3}{\sqrt{18-9}}=-1$