Question

Which of the following curves cut the parabola $y^2=4ax$ at right angles

• A. $x^2+y^2=a^2$
• B. $y=e^{-\frac{x}{2a}}$
• C. $y=ax$
• D. $x^2=4ay$
• E. B and D

Answer 1

The correct option is D

$x^2=4ay$

The explanation for the correct option

Option D: $x^2=4ay$ is the general equation of a parabola symmetric to the y-axis opening upwards with the tangent perpendicular to the axis of symmetry.

The intersection is at the right angle.

$y^2=4ax$ is the general equation of a parabola symmetric to the x-axis opening rightwards with the tangent perpendicular to the axis of symmetry.

The tangent of the parabola $x^2=4ay$ at $\left(0,0\right)$ is the X-axis and the tangent of the parabola $y^2=4ax$at $\left(0,0\right)$ is the Y-axis.

Since X and Y axes are perpendicular to each other, The parabolas cut each other at right angle.

Hence, the correct option is D

The explanation for incorrect options

Option A: $x^2+y^2=a^2$ is a circle with center as $\left(0,0\right)$ and radius $a$. The intersection is not at the right angle.

Option B: $y=e^{-\frac{x}{2a}}$ is an exponential function. The intersection is not at the right angle.

Option C: $y=ax$is a straight line. The intersection is not at the right angle.

Hence, the correct option is D