Question:
If \(\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is
If \(\frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is
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Differentiate the ellipse equation implicitly twice and simplify using the ellipse equation.
Updated On: Nov 13, 2026
- \(\frac{-b^{4}}{a}\)
- \(\frac{\mathrm{b}^{4}}{\mathrm{a}^{2}}\)
- \(\frac{-\mathrm{b}^{4}}{y^{3}}\)
- \(\frac{-b^{4}}{a^{2}\cdot y^{3}}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Differentiate \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) implicitly with respect to \(x\).
Step 2: First derivative
\(\dfrac{2x}{a^2} + \dfrac{2y}{b^2}\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{b^2 x}{a^2 y}\).
Step 3: Second derivative
\(\dfrac{d^2y}{dx^2} = -\dfrac{b^2}{a^2}\cdot\dfrac{y - x\,y'}{y^2} = -\dfrac{b^2}{a^2}\cdot\dfrac{y + \frac{b^2x^2}{a^2y}}{y^2}\)
\(= -\dfrac{b^2}{a^2}\cdot\dfrac{a^2y^2 + b^2x^2}{a^2y^3}\)
Step 4: Use the ellipse equation
Since \(b^2x^2 + a^2y^2 = a^2b^2\):
\(\dfrac{d^2y}{dx^2} = -\dfrac{b^2}{a^2}\cdot\dfrac{a^2b^2}{a^2y^3} = -\dfrac{b^4}{a^2y^3}\).
Final Answer:
The second derivative is -b^4 divided by (a^2 y^3). \[ \boxed{\text{(D) }\dfrac{-b^{4}}{a^{2}\cdot y^{3}}} \]
Differentiate \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) implicitly with respect to \(x\).
Step 2: First derivative
\(\dfrac{2x}{a^2} + \dfrac{2y}{b^2}\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{b^2 x}{a^2 y}\).
Step 3: Second derivative
\(\dfrac{d^2y}{dx^2} = -\dfrac{b^2}{a^2}\cdot\dfrac{y - x\,y'}{y^2} = -\dfrac{b^2}{a^2}\cdot\dfrac{y + \frac{b^2x^2}{a^2y}}{y^2}\)
\(= -\dfrac{b^2}{a^2}\cdot\dfrac{a^2y^2 + b^2x^2}{a^2y^3}\)
Step 4: Use the ellipse equation
Since \(b^2x^2 + a^2y^2 = a^2b^2\):
\(\dfrac{d^2y}{dx^2} = -\dfrac{b^2}{a^2}\cdot\dfrac{a^2b^2}{a^2y^3} = -\dfrac{b^4}{a^2y^3}\).
Final Answer:
The second derivative is -b^4 divided by (a^2 y^3). \[ \boxed{\text{(D) }\dfrac{-b^{4}}{a^{2}\cdot y^{3}}} \]
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