Finding the common tangents to the parabola $y^2=15x$ and the circle $x^2+y^2=16$

Question

I'm fine with part i, need help with part ii. And given that the normal to the tangent is y=-1/m.

Answer 1

By your work the equation of the tangent it's: $$mx-y+\frac{15}{4m}=0$$ and since the distance from the center of the circle and the tangent is equal to the radius of the circle, we obtain: $$\frac{\left|m\cdot0-0+\frac{15}{4m}\right|}{\sqrt{m^2+1}}=4,$$ which gives $m=\frac{3}{4}$ or $m=-\frac{3}{4}.$

Can you end it now?

Answer 2

Hint

Once you already know that $y=mx+15/4m$ is tangent to the parabola. It is just necessary to put it tangent to the circle. On way to do that is replace that value of $y$ into the circle equation:

$$x^2+\left(mx+\frac{15}{4m}\right)^2=16$$

It will give you a quadradic equation and you have to force it to have only one solution.

Can you finish?