Question

Find the points on the curve $y=x^3$, the tangents at which are inclined at an angle of ${60}^{\circ }$ to x-axis.

• A. $x=\pm \frac{1}{\sqrt{\sqrt{3}}},y=\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{\sqrt{3}}}$.
• B. $x=\frac{1}{\sqrt{\sqrt{3}}},y=\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{\sqrt{3}}}$.
• C. $x=\pm \frac{1}{\sqrt{\sqrt{3}}},y=\pm \frac{1}{\sqrt{3}}.\frac{1}{\sqrt{\sqrt{3}}}$.
• D. $x=-\frac{1}{\sqrt{\sqrt{3}}},y=\pm \frac{1}{\sqrt{3}}.\frac{1}{\sqrt{\sqrt{3}}}$.

Answer 1

The correct option is A $x=\pm \frac{1}{\sqrt{\sqrt{3}}},y=\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{\sqrt{3}}}$.

Hence slope of tangents
$=tan{60}^0$
$=\sqrt{3}$
$=\frac{dy}{dx}$
Hence
$\frac{dy}{dx}$
$=3x^2$
$=\sqrt{3}$
Or
$x^2=\frac{1}{\sqrt{3}}$
Hence
$x=\pm \frac{1}{\sqrt{\sqrt{3}}}$.
Thus y
$=\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{\sqrt{3}}}$.