1. **Problem Statement:** Find the distance of the point $P$ from the origin, where $P$ is the intersection of the common tangents to the parabola $x^2 = 4y$ and the circle $x^2 + y^2 = 4$. Also, explain why $P$ lies on the $y$-axis and not in the 3rd or 4th quadrant. 2. **Equations of curves:** - Parabola: $$x^2 = 4y$$ - Circle: $$x^2 + y^2 = 4$$ 3. **Equation of tangent to parabola:** The tangent to the parabola $x^2 = 4ay$ can be written as $$y = mx + a/m$$ where $a=1$ (since $4a=4$), so $$y = mx + \frac{1}{m}$$ 4. **Condition for tangent to the circle:** Substitute $y = mx + \frac{1}{m}$ into the circle equation: $$x^2 + \left(mx + \frac{1}{m}\right)^2 = 4$$ Expanding: $$x^2 + m^2x^2 + 2x + \frac{1}{m^2} = 4$$ $$x^2(1 + m^2) + 2x + \frac{1}{m^2} - 4 = 0$$ 5. **Tangency condition:** For the line to be tangent to the circle, the quadratic in $x$ must have a discriminant of zero: $$\Delta = 2^2 - 4(1 + m^2)\left(\frac{1}{m^2} - 4\right) = 0$$ Simplify: $$4 - 4(1 + m^2)\left(\frac{1}{m^2} - 4\right) = 0$$ Divide both sides by 4: $$1 - (1 + m^2)\left(\frac{1}{m^2} - 4\right) = 0$$ 6. **Simplify the expression:** $$1 = (1 + m^2)\left(\frac{1}{m^2} - 4\right) = (1 + m^2)\frac{1}{m^2} - 4(1 + m^2) = \frac{1 + m^2}{m^2} - 4 - 4m^2$$ Multiply both sides by $m^2$: $$m^2 = 1 + m^2 - 4m^2 - 4m^4$$ Simplify: $$m^2 = 1 + m^2 - 4m^2 - 4m^4$$ $$0 = 1 - 4m^2 - 4m^4$$ Rearranged: $$4m^4 + 4m^2 - 1 = 0$$ 7. **Solve quadratic in $m^2$:** Let $t = m^2$, then $$4t^2 + 4t - 1 = 0$$ Using quadratic formula: $$t = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm \sqrt{32}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} = \frac{-1 \pm \sqrt{2}}{2}$$ 8. **Valid solutions for $m^2$:** Since $m^2 \geq 0$, only $$m^2 = \frac{-1 + \sqrt{2}}{2}$$ is valid. 9. **Find $m$ and corresponding tangents:** $$m = \pm \sqrt{\frac{-1 + \sqrt{2}}{2}}$$ The tangents are: $$y = mx + \frac{1}{m}$$ 10. **Find intersection point $P$ of the two tangents:** Set the two tangent lines equal: $$m x + \frac{1}{m} = -m x - \frac{1}{m}$$ $$2 m x = - \frac{2}{m}$$ $$x = -\frac{1}{m^2}$$ Substitute back to find $y$: $$y = m x + \frac{1}{m} = m \left(-\frac{1}{m^2}\right) + \frac{1}{m} = -\frac{1}{m} + \frac{1}{m} = 0$$ So, the intersection point $P$ is: $$\left(-\frac{1}{m^2}, 0\right)$$ 11. **Check the sign of $x$ coordinate:** Since $m^2 = \frac{-1 + \sqrt{2}}{2} > 0$, then $$x = -\frac{1}{m^2} < 0$$ So $P$ lies on the negative $x$-axis, not on the $y$-axis. 12. **Distance from origin:** $$d = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{1}{m^2}\right)^2 + 0^2} = \frac{1}{m^2}$$ Substitute $m^2$: $$d = \frac{1}{\frac{-1 + \sqrt{2}}{2}} = \frac{2}{-1 + \sqrt{2}} = \frac{2}{\sqrt{2} - 1}$$ Rationalize denominator: $$d = \frac{2}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = 2(\sqrt{2} + 1) = 2\sqrt{2} + 2$$ 13. **Why $P$ lies on the $x$-axis and not in 3rd or 4th quadrant:** The intersection $P$ is the point where the two common tangents meet. From step 10, $P$ has $y=0$, so it lies on the $x$-axis. Since $x < 0$, it lies on the negative $x$-axis (2nd quadrant boundary), not in the 3rd or 4th quadrant. The problem's assumption that $P$ lies on the $y$-axis is incorrect based on this calculation. **Final answer:** The distance of $P$ from the origin is $$2\sqrt{2} + 2$$ and $P$ lies on the negative $x$-axis, not on the $y$-axis.