Question

The point on the curve $y=16-x^2,x\in \left(-11\right]$, where tangent is parallel to x-axis is .

• A. (16, 0)
• B. (0, 16)
• C. (4, 0)
• D. (0, 4)

Answer 1

The correct option is B (0, 16)

The equation of the curve is$y=16-x^2,x\in \left(-11\right]$. Let P(a, b) be the point where tangent to the given curve is parallel to x-axis. Then, ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_P=-2a\therefore {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_P=0\Rightarrow -2a=0\Rightarrow a=0\because P\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{given}\mathrm{curve}\therefore b=16-a^2\Rightarrow b=16.$
Hence, (0, 16) is the required point.