Question

The differential equation of an ellipse whose major axis is twice its minor axis, is

• A. $x + 4y \frac{dy}{dx} = 0$
• B. $x - 4y \frac{dy}{dx} = 0$
• C. $x + 2y \frac{dy}{dx} = 0$
• D. None of these

Hint:

Whenever you are asked to form a differential equation from a geometric condition, first translate that condition into an algebraic equation to eliminate as many arbitrary constants as possible before differentiating.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to formulate the differential equation for a specific family of ellipses where the length of the major axis is exactly double the length of the minor axis.

Step 2: Key Formula or Approach:
The standard equation of an ellipse centered at the origin is:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $2a$ is the major axis and $2b$ is the minor axis. After applying the given condition to eliminate one parameter, we differentiate the equation implicitly with respect to $x$ to find the differential equation.

Step 3: Detailed Explanation:
The problem states that the major axis is twice the minor axis:
$$2a = 2(2b) \implies a = 2b$$ Substitute $a = 2b$ into the standard equation of the ellipse:
$$\frac{x^2}{(2b)^2} + \frac{y^2}{b^2} = 1$$ $$\frac{x^2}{4b^2} + \frac{y^2}{b^2} = 1$$ Multiply the entire equation by $4b^2$ to simplify:
$$x^2 + 4y^2 = 4b^2$$ Now, differentiate both sides implicitly with respect to $x$. Remember that $4b^2$ is a constant, so its derivative is zero:
$$2x + 4\left(2y \frac{dy}{dx}\right) = 0$$ $$2x + 8y \frac{dy}{dx} = 0$$ Divide the entire equation by 2:
$$x + 4y \frac{dy}{dx} = 0$$

Step 4: Final Answer:
The differential equation is $x + 4y \frac{dy}{dx} = 0$, matching option (A).