Skip to content

Parabola

Mediummathematics

Consider the curve traced by the parametric pair x = 2t^2 and y = 4t as the parameter t varies over all real numbers; identify the Cartesian equation.

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

More coordinate geometry Practice Questions

EasyJEE_ADVANCED

For the parabola described by y^2 = 12x, determine the coordinates of its focus together with the eq...

View Solution→
MediumJEE_ADVANCED

A chord of the parabola y^2 = 8x passes through its focus and is perpendicular to the axis; what is ...

View Solution→
EasyJEE_MAIN

The standard equation of a parabola with vertex at origin and focus on positive x-axis is

View Solution→
EasyJEE_MAIN

The focus of the parabola y² = 8x is

View Solution→
EasyJEE_ADVANCED

Statement: the lines 2x - 3y + 5 = 0 and 3x + 2y - 7 = 0 are mutually perpendicular; which option co...

View Solution→
EasyJEE_ADVANCED

A point divides the line segment joining (2, -3) and (8, 9) internally in the ratio 1 : 2; what are ...

View Solution→
HardJEE_ADVANCED

A point moves so that the sum of squares of its distances from the two points (3, 0) and (-3, 0) sta...

View Solution→
HardJEE_ADVANCED

A rectangular hyperbola is described by the equation xy = 8 in the coordinate plane; what is the val...

View Solution→
HardJEE_ADVANCED

An ellipse has foci separated by a distance of 6 units and a sum of focal radii equal to 10 for ever...

View Solution→
EasyJEE_ADVANCED

The straight line 4x + 3y - 24 = 0 forms a triangle with the two coordinate axes; what is the area o...

View Solution→
MediumJEE_ADVANCED

From the external point (5, 1) a tangent is drawn to the circle x^2 + y^2 = 9; what is the length of...

View Solution→
MediumJEE_ADVANCED

A moving point keeps its distance from the point (4, 0) always equal to its distance from the vertic...

View Solution→

Browse all coordinate geometry questions →

About This Question

Subject
mathematics
Chapter
coordinate geometry
Topic
parabola
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillparabolaparametric formparameter eliminationCartesian equation

Solution

Correct Answer:

Eliminating the parameter from a parametric pair recovers the Cartesian relation, and recognizing the resulting form lets us classify the conic. From y = 4t we get t = y/4, and substituting into x = 2t^2 gives x = 2(y/4)^2 = 2 · y^2/16 = y^2/8. Multiplying through, y^2 = 8x, a right-opening parabola. Option y^2 = 4x would arise from x = t^2 rather than 2t^2. Option x^2 = 8y wrongly treats x as the squared parameter expression. Option y^2 = 16x comes from mis-squaring y = 4t as if x = t^2. This is the standard JEE Advanced parametric-to-Cartesian elimination, a recurring conic technique. Plausibility check: comparing y^2 = 8x with y^2 = 4ax gives a = 2, and sample t = 1 yields the point (2, 4), which satisfies 16 = 8·2, confirming the curve passes through the generated points consistently.

This medium difficulty mathematics question is from the chapter coordinate geometry, covering the topic of parabola. It appeared in the 2025 exam.

Looking for more practice? Explore all mathematics questions or browse coordinate geometry questions on RankGuru.