Question

Let $y^{2}=8x$ be the equation of a parabola. Which one of the following is an arbitrary point on the parabola?

• A. $(2t, 4t^2)$
• B. $(2t^2, 4t^2)$
• C. $(2t^2, 2t^2)$
• D. $(2t, 2t^2)$
• E. $(2t^2, 4t)$

Hint:

Always identify the value of '$a$' first before writing the parametric coordinates.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
An "arbitrary point" on a curve can be represented by its parametric coordinates. Parametric equations express the coordinates x and y as functions of a single parameter, typically 't'. For a standard parabola, there is a standard parametric form.
Step 2: Key Formula or Approach:
The standard equation of a right-opening parabola is \( y^2 = 4ax \).
The parametric coordinates for any point on this parabola are given by:
\[ x = at^2 \] \[ y = 2at \] where 't' is the parameter.
Step 3: Detailed Explanation:
First, we need to find the value of 'a' for the given parabola \( y^2 = 8x \).
Compare \( y^2 = 8x \) with the standard form \( y^2 = 4ax \):
\[ 4a = 8 \] \[ a = \frac{8}{4} = 2 \] Now that we have \( a=2 \), we can substitute this value into the standard parametric equations:
For the x-coordinate:
\[ x = at^2 = 2t^2 \] For the y-coordinate:
\[ y = 2at = 2(2)t = 4t \] So, the arbitrary point on the parabola \( y^2 = 8x \) is \( (2t^2, 4t) \).
We can verify this by substituting these coordinates back into the parabola's equation:
\[ y^2 = (4t)^2 = 16t^2 \] \[ 8x = 8(2t^2) = 16t^2 \] Since \( y^2 = 8x \) holds true, the point is on the parabola.
Step 4: Final Answer:
The arbitrary point on the parabola is \( (2t^2, 4t), t \in \mathbb{R} \).