Question

If $L$ is the normal drawn to the parabola $y^2 = 8x$ at the point $t = \frac{1}{\sqrt{2}}$, then the foot of the perpendicular drawn from the focus of the parabola onto the normal $L$ is:

• A. $(3, 2)$
• B. $\left(5, \sqrt{2}\right)$
• C. $\left(0, \sqrt{2}\right)$
• D. $\left(3, \sqrt{2}\right)$

Hint:

To find the foot of perpendicular from a point to a line, parametrize the foot using the normal vector direction and solve using the line equation.

Answer 1

The Correct Option is D

Step 1: Parametric point on parabola
For $y^2 = 4ax$, parametric point at $t$ is $(a t^2, 2 a t)$. Given $y^2 = 8x \Rightarrow 4a = 8 \Rightarrow a = 2$.
At $t = \frac{1}{\sqrt{2}}$: \[ x = 2 \times \left(\frac{1}{\sqrt{2}}\right)^2 = 1, \quad y = 4 \times \frac{1}{\sqrt{2}} = 2 \sqrt{2}. \] Step 2: Equation of normal
Normal to parabola at $t$: \[ y + t x = 2 a t + a t^3. \] Substitute $a=2$, $t = \frac{1}{\sqrt{2}}$: \[ y + \frac{1}{\sqrt{2}} x = 4 \times \frac{1}{\sqrt{2}} + 2 \times \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{4}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}. \] Multiply by $\sqrt{2}$: \[ \sqrt{2} y + x = 5. \] Step 3: Coordinates of focus
Focus of parabola is $(a, 0) = (2, 0)$. Step 4: Foot of perpendicular from focus to normal
Line: $x + \sqrt{2} y - 5 = 0$.
Point: $(2,0)$. Let foot be $(x_f, y_f)$. The vector from $(2,0)$ to foot is perpendicular to the line, so: \[ \frac{x_f - 2}{1} = \frac{y_f - 0}{\sqrt{2}} = -\lambda. \] So, \[ x_f = 2 - \lambda, \quad y_f = -\lambda \sqrt{2}. \] Substitute into line: \[ (2 - \lambda) + \sqrt{2}(-\lambda \sqrt{2}) - 5 = 0, \] \[ 2 - \lambda - 2 \lambda - 5 = 0, \] \[ -3 \lambda - 3 = 0 \implies \lambda = -1. \] Therefore, \[ x_f = 3, \quad y_f = \sqrt{2}. \]