Question

Set up an equation of a tangent to the graph of the following function.
The tangent to the curve $y=4-x^2$ makes an angle of ${75}^{\circ }$ with the X axis. Find the coordinates of the point of tangency.

Answer 1

$y=4-x^2$

$\frac{dy}{dx}=-2x=\tan {75}^0$

$-2x=\tan {75}^0$

If we substitute $x=-\frac{\sqrt{3}+1}{4\sqrt{2}}$

$y=4-\left(\frac{4+2\sqrt{3}}{32}\right)=\frac{128-2\sqrt{3}}{32}=y$

The equation of tangent is $\frac{\sqrt{3}+1}{2\sqrt{2}}(x+\frac{\sqrt{3}+1}{4\sqrt{2}})=y-\frac{128-2\sqrt{3}}{32}$

$\frac{\sqrt{3}+1}{2\sqrt{2}}x-y+\frac{4+2\sqrt{3}}{16}+\frac{128-2\sqrt{3}}{32}=0$

$\frac{\sqrt{3}+1}{2\sqrt{2}}x-y+\frac{136+2\sqrt{3}}{32}=0$