Question:

Step 1: Determine the general equation of a tangent to the parabola. The equation of the parabola is $y^2 = 4x$. This is in the form $y^2 = 4ax$, where $a=1$. The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is given by: $$y = mx + \frac{a}{m}$$ Substituting $a=1$, the tangent equation is: $$y = mx + \frac{1}{m}$$ Step 2: Use the external point to find the slopes of the tangents. The tangents are drawn from the point $(-1, 2)$. Substitute this point into the tangent equation: $$2 = m(-1) + \frac{1}{m}$$ $$2 = -m + \frac{1}{m}$$ Multiply by $m$ (assuming $m \neq 0$, which is true for tangents to $y^2=4x$ from an external point): $$2m = -m^2 + 1$$ Rearrange into a quadratic equation: $$m^2 + 2m - 1 = 0$$ Step 3: Solve for the slopes of the two tangents. Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$m = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}$$ $$m = \frac{-2 \pm \sqrt{4 + 4}}{2}$$ $$m = \frac{-2 \pm \sqrt{8}}{2}$$ $$m = \frac{-2 \pm 2\sqrt{2}}{2}$$ $$m = -1 \pm \sqrt{2}$$ Let the two slopes be $m_1 = -1 + \sqrt{2}$ and $m_2 = -1 - \sqrt{2}$. Step 4: Determine the equations of the two tangents. For $m_1 = -1 + \sqrt{2}$: The term $\frac{1}{m_1} = \frac{1}{-1 + \sqrt{2}} = \frac{1}{\sqrt{2} - 1}$. Rationalize the denominator: $$\frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1$$ So, the first tangent equation is: $$y_1 = (-1 + \sqrt{2})x + (1 + \sqrt{2})$$ For $m_2 = -1 - \sqrt{2}$: The term $\frac{1}{m_2} = \frac{1}{-1 - \sqrt{2}} = \frac{1}{-(1 + \sqrt{2})}$. Rationalize the denominator: $$\frac{1}{-(1 + \sqrt{2})} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{-(1 - 2)} = \frac{1 - \sqrt{2}}{-(-1)} = 1 - \sqrt{2}$$ So, the second tangent equation is: $$y_2 = (-1 - \sqrt{2})x + (1 - \sqrt{2})$$ Step 5: Find the points where the tangents intercept the line $x = 2$. Substitute $x=2$ into the tangent equations: For $y_1$: $$Y_1 = (-1 + \sqrt{2})(2) + (1 + \sqrt{2})$$ $$Y_1 = -2 + 2\sqrt{2} + 1 + \sqrt{2}$$ $$Y_1 = -1 + 3\sqrt{2}$$ For $y_2$: $$Y_2 = (-1 - \sqrt{2})(2) + (1 - \sqrt{2})$$ $$Y_2 = -2 - 2\sqrt{2} + 1 - \sqrt{2}$$ $$Y_2 = -1 - 3\sqrt{2}$$ The two points of intersection on the line $x=2$ are $(2, -1 + 3\sqrt{2})$ and $(2, -1 - 3\sqrt{2})$. Step 6: Calculate the length intercepted on the line $x = 2$. The length intercepted is the distance between these two points. Since their x-coordinates are the same, the distance is the absolute difference of their y-coordinates: $$\text{Length} = |Y_1 - Y_2|$$ $$\text{Length} = |(-1 + 3\sqrt{2}) - (-1 - 3\sqrt{2})|$$ $$\text{Length} = |-1 + 3\sqrt{2} + 1 + 3\sqrt{2}|$$ $$\text{Length} = |6\sqrt{2}|$$ $$\text{Length} = 6\sqrt{2}$$