Question

Two parabola $y^2=4a(x-{\lambda }_1),$ and $x^2=4a(y-{\lambda }_2)$ always touch each other, where ${\lambda }_1$ and ${\lambda }_2$ being variable parameters. Then their points of contact lie on a

• A. Circle
• B. Hyperbola
• C. Ellipse
• D. Parabola

Answer 1

The correct option is B

Hyperbola

Since the parabolas touch each other, they will have the common tangent.

Finding $\frac{dy}{dx}$:

$y^2=4a(x-{\lambda }_1)$

Differentiating with respect to $x$, we get

$2y\frac{dy}{dx}=4a$ ... (1)

$\Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}$

Differentiating $x^2=4a(y-{\lambda }_2)$ with respect to $y$, we get

$2x\frac{dx}{dy}=4a$

$\Rightarrow \frac{dy}{dx}=\frac{2x}{4a}=\frac{x}{2a}$ ... (2)

Equating equations (1) and (2), we get

$\frac{2a}{y}=\frac{x}{2a}$

$\Rightarrow xy=4a^2$ which is a hyperbola.

Hence, their points of contact lie on a hyperbola.