Question

[Key Concept: A common tangent to $|y| = x^{2}$ and the circle $x^{2} + (y{- 2)}^{2} = 4$: impose tangency to the parabola and the distance-to-centre condition simultaneously.]

Eqn. of tangent to $y = x^{2}$ is

$$tx = y + \frac{1}{4}t^{2}$$

⟂ at from centre $(0,2) = 2$

$${\Rightarrow \frac{\left| \frac{1}{4}t^{2} + 2 \right|}{\sqrt{1 + t^{2}}} = 2 }{\Rightarrow \left( t^{2} + 8 \right)^{2} = 64\left( 1 + t^{2} \right) }{\Rightarrow t^{4} - 48t^{2} = 0 }{\Rightarrow \ t = 0, \pm 4\sqrt{3} }$$Eqn. of tangent is $\pm 4\sqrt{3}x = y + 12$

Eqn. of tangent to $y = - x^{2}$ is

$$tx = y - \frac{1}{4}t^{2} $$⟂ at from $(0,2) = 2$

$$\Rightarrow \frac{\left( \frac{1}{4}t^{2} - 2 \right)}{\sqrt{1 + t^{2}}} = 2$$

$${\Rightarrow t^{4} - 80t^{2} = 0 }{\Rightarrow \ t = \pm 4\sqrt{5} }$$∴ Equation of tangents are

$$\pm 4\sqrt{5}x = y - 20$$