Question:

**Step 1:** Let $P$ be a point on the parabola $S_1: y^2 = 4ax$. The parametric coordinates for $P$ are $(at^2, 2at)$. **Step 2:** The equation of the chord of contact from $P(at^2, 2at)$ to the parabola $S_2: y^2 = -4ax$ is given by $T=0$. $$y(2at) = -4a \left(\frac{x + at^2}{2}\right)$$ $$2aty = -2a(x + at^2)$$ Dividing by $2a$ (assuming $a \neq 0$): $$ty = -(x + at^2)$$ $$x + ty + at^2 = 0$$ This is the equation of the line $QR$. **Step 3:** To determine if the line $QR: x + ty + at^2 = 0$ is tangent to $S_1: y^2 = 4ax$, substitute $x = -ty - at^2$ into the equation of $S_1$: $$y^2 = 4a(-ty - at^2)$$ $$y^2 = -4aty - 4a^2t^2$$ $$y^2 + 4aty + 4a^2t^2 = 0$$ This is a quadratic equation in $y$. It can be factored as: $$(y + 2at)^2 = 0$$ Since this equation yields a repeated root for $y$, the line $QR$ is tangent to $S_1$. **Step 4:** The equation of the normal to the parabola $S_1: y^2 = 4ax$ at a point $A(at^2, 2at)$ is given by: $$y = -tx + 2at + at^3$$ **Step 5:** For this normal line to be tangent to the parabola $S_2: y^2 = -4ax$, we use the condition of tangency. For a parabola $y^2 = 4Ax$, a line $y = mx + c$ is tangent if $c = A/m$. For $S_2: y^2 = -4ax$, we have $A = -a$. The normal line has slope $m = -t$ and y-intercept $c = 2at + at^3$. Applying the tangency condition $c = \frac{-a}{m}$: $$2at + at^3 = \frac{-a}{-t}$$ $$2at + at^3 = \frac{a}{t}$$ Assuming $a \neq 0$ and $t \neq 0$, we can divide by $a$ and multiply by $t$: $$t(2t + t^3) = 1$$ $$2t^2 + t^4 = 1$$ $$t^4 + 2t^2 - 1 = 0$$ **Step 6:** Solve the quadratic equation in $t^2$: Let $u = t^2$. Then $u^2 + 2u - 1 = 0$. Using the quadratic formula: $$u = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}$$ $$u = \frac{-2 \pm \sqrt{4 + 4}}{2}$$ $$u = \frac{-2 \pm \sqrt{8}}{2}$$ $$u = \frac{-2 \pm 2\sqrt{2}}{2}$$ $$u = -1 \pm \sqrt{2}$$ Since $t^2 = u$ must be positive, we take the positive root: $$t^2 = \sqrt{2} - 1$$