Question

$AB$ is a chord of a parabola $y^2 = 4ax$, $(a > 0)$ with vertex $A$, $BC$ is drawn perpendicular to $AB$ meeting the axis at $C$. The projection of $BC$ on the axis of the parabola is

• A. a unit
• B. 2a unit
• C. 8a unit
• D. 4a unit

Answer 1

The Correct Option is D

Given: - Parabola: \( y^2 = 4ax \), with vertex at the origin \( A = (0, 0) \), and axis along the x-axis - \( AB \) is a chord with vertex at one endpoint - \( BC \perp AB \), and intersects the axis of the parabola at point \( C \) - We are to find the projection of \( BC \) on the axis (x-axis) Step 1: Assume coordinates Let \( B = (x_1, y_1) \) lie on the parabola, so: \[ y_1^2 = 4a x_1 \] Let \( A = (0, 0) \) (vertex) So, vector \( \overrightarrow{AB} = (x_1, y_1) \) Since \( BC \perp AB \), the direction vector of \( BC \) is perpendicular to \( AB \) Therefore, direction vector of \( BC \) is: \[ (-y_1, x_1) \] Let us find the x-component (projection on x-axis) of \( \overrightarrow{BC} \). Since the direction vector of \( BC \) is \( (-y_1, x_1) \), its projection on the x-axis is the magnitude of the x-component: \[ | -y_1 | = |y_1| \] Now since \( y_1^2 = 4a x_1 \), we get: \[ y_1 = \sqrt{4a x_1} \] So projection = \( |y_1| = \sqrt{4a x_1} \) But we don't know \( x_1 \). Instead, let’s use geometry to get exact value. Step 2: Use geometry — Projection of perpendicular from B onto axis Let’s choose point \( B = (4a, 4a) \) Check: does it lie on parabola? \[ y^2 = (4a)^2 = 16a^2,\quad 4ax = 4a \cdot 4a = 16a^2 \Rightarrow \text{Yes} \] So, \( A = (0, 0) \), \( B = (4a, 4a) \) Then vector \( \overrightarrow{AB} = (4a, 4a) \) Direction vector of \( BC \) is perpendicular to this, so it's \( (-4a, 4a) \) Now find where this line intersects the x-axis: Parametrize the line through \( B \) in direction \( (-4a, 4a) \): \[ x = 4a - 4a t,\quad y = 4a + 4a t \] For x-axis intersection, set \( y = 0 \): \[ 0 = 4a + 4a t \Rightarrow t = -1 \Rightarrow x = 4a - 4a(-1) = 8a \] So, point \( C = (8a, 0) \) Then \( BC \) connects \( B = (4a, 4a) \) and \( C = (8a, 0) \) Projection of \( BC \) on x-axis = difference in x-coordinates = \[ 8a - 4a = 4a \] Final Answer: \[ \boxed{4a\ \text{unit}} \]