What is the differential equation of all parabolas of the type y^{2}=4a(x-b)?

  1. A. \frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=0
  2. B. \frac{d^{2}y}{dx^{2}}+x^{2}\left(\frac{dy}{dx}\right)^{2}=0
  3. C. y^{2}\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=0
  4. D. y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=0

Correct Answer: D. y\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}=0

Explanation

The equation is y^2 = 4a(x-b). Differentiating with respect to x gives 2y \frac{dy}{dx} = 4a, which simplifies to y \frac{dy}{dx} = 2a. This eliminates the arbitrary constant b. To eliminate a, differentiate again with respect to x using the product rule: y \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{dy}{dx} = 0, yielding y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0.

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