If x = a t^2 and y = 2 a t for a real parameter t, which relation between x, y, and a is always true for all permissible values of t?

Introduction / Context:
This question involves algebraic manipulation of parametric equations. You are given x and y in terms of a parameter t and a constant a, and you must eliminate t to find a direct relationship between x, y, and a. Such problems are common when studying coordinate geometry, especially in deriving standard forms of curves like parabolas from parametric definitions.

Given Data / Assumptions:

• x = a t^2.
• y = 2 a t.
• a is a constant, and t is a real parameter.
• We want a relation that holds for all values of t and does not include t explicitly.

Concept / Approach:
To eliminate the parameter t, we express t in terms of y and a from one equation and substitute into the other. From y = 2 a t, we can write t = y / (2a). Substituting into x = a t^2 then yields x in terms of y and a only. After simplification, we rearrange to match one of the given options. This is the standard technique for deriving Cartesian equations from parametric ones.

Step-by-Step Solution:
Given y = 2 a t, solve for t: t = y / (2a).Substitute this expression for t into x = a t^2.Compute t^2: t^2 = (y / (2a))^2 = y^2 / (4a^2).Therefore, x = a * (y^2 / (4a^2)) = y^2 / (4a).Multiply both sides by 4a to clear the denominator: 4a x = y^2.Rewriting, we obtain the relationship y^2 = 4 a x.

Verification / Alternative check:
We can verify the relation by choosing a convenient numerical value for t and checking whether y^2 equals 4 a x. For example, let a = 1 and t = 2. Then x = 1 * 2^2 = 4 and y = 2 * 1 * 2 = 4. Now y^2 = 16 and 4 a x = 4 * 1 * 4 = 16, so the relation y^2 = 4 a x holds. Trying other values for t will give the same confirmation, indicating that this equation is identically true for all choices of t and a (with a non zero).

Why Other Options Are Wrong:

• x^2 = 4 a y would suggest squaring x instead of y and does not follow from substituting t.
• x^2 + y^2 = a^2 and x^2 − y^2 = a^2 resemble circle or hyperbola forms, which are not supported by the given parametric definitions.
• xy = 2a is not consistent with the actual dependence on t; substituting x and y shows that xy depends on t^3, not a constant.

Common Pitfalls:

• Solving for t incorrectly from y = 2 a t, leading to errors when substituting into x.
• Mixing up which variable is squared when forming the final relation.
• Assuming a circle or hyperbola shape without checking the algebraic form obtained after elimination.

Final Answer:
y^2 = 4ax