Question

The differential equation of all parabolas, whose axes are parallel to Y-axis, is

• A. $y_3=1$
• B. $y_3=0$
• C. $y_3=-1$
• D. $yy_3+y_1=0$

Hint:

Logic Tip: The general equation can also be written in the form $y = Ax^2 + Bx + C$. Taking the first derivative gives $y_1 = 2Ax + B$, the second gives $y_2 = 2A$, and the third derivative immediately eliminates the final constant to give $y_3 = 0$.

Answer 1

The Correct Option is B

Concept:
The order of the differential equation of a family of curves is equal to the number of independent arbitrary constants in its general equation. To find the differential equation, we differentiate the general equation as many times as there are arbitrary constants to eliminate them.
Step 1: Write the general equation of the family of parabolas.
The general equation of a parabola with its axis parallel to the Y-axis has a vertex at $(h,k)$ and can be written as: $$(x-h)^2 = 4b(y-k)$$ This equation has three arbitrary constants: $h$, $k$, and $b$. Therefore, we expect a 3rd-order differential equation.
Step 2: Differentiate successively to eliminate the constants.
Differentiate once with respect to $x$: $$2(x-h) = 4b\left(\frac{dy}{dx}\right) \implies 2(x-h) = 4by_1$$ Differentiate a second time with respect to $x$: $$2 = 4b\left(\frac{d^2y}{dx^2}\right) \implies 2 = 4by_2$$ Differentiate a third time with respect to $x$: $$0 = 4b\left(\frac{d^3y}{dx^3}\right) \implies 0 = 4by_3$$ Since $4b$ is not zero (otherwise it wouldn't be a parabola), we divide by $4b$ to get: $$y_3 = 0$$