Question

The differential equation of a family of hyperbolas whose axes are parallel to coordinate axes, centres lie on the line $y=2x$ and eccentricity is $\sqrt{3}$ is

• A. $(2x-y)y_2 + y_1^2 - 2y_1 = y_1^3+2$
• B. $(y-2x)y_2 + y_1^2 + 2y_1 = y_1^3+2$
• C. $(y-2x)y_2 - y_1^2 + 2y_1 = y_1^3-2$
• D. $(y+2x)y_2 + y_1^2 + 2y_1 = y_1^3-2$

Hint:

Forming differential equations involves eliminating arbitrary constants. If there are $n$ constants, you generally need to differentiate $n$ times. The process often involves a lot of algebraic manipulation. Keep the expressions for the derivatives and try to solve for the constants (or expressions involving them) to substitute them into higher-order derivative equations.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
We need to form a differential equation for a given family of curves. This involves writing the general equation of the family, which will contain some arbitrary constants (parameters), and then eliminating these constants by differentiating the equation a sufficient number of times.
Step 2: Key Formula or Approach
1. Write the general equation of a hyperbola with axes parallel to the coordinate axes and center $(h,k)$. 2. Incorporate the given conditions: center lies on $y=2x$, so $k=2h$. Eccentricity $e=\sqrt{3}$. 3. The eccentricity condition relates the semi-axes $a$ and $b$: $e^2 = 1+\frac{b^2}{a^2}$ or $e^2 = 1+\frac{a^2}{b^2}$. 4. Differentiate the general equation with respect to $x$ to eliminate the parameters.
Step 3: Detailed Explanation
1. General Equation and Conditions: Let the center be $(h,k)$. The equation of the hyperbola is $\frac{(x-h)^2}{A} - \frac{(y-k)^2}{B} = 1$. (We use A and B to avoid confusion with semi-axes a,b). Given $k=2h$. Eccentricity $e=\sqrt{3} \implies e^2=3$. $e^2 = 1 + \frac{B}{A} \implies 3 = 1 + \frac{B}{A} \implies \frac{B}{A}=2 \implies B=2A$. The equation becomes $\frac{(x-h)^2}{A} - \frac{(y-2h)^2}{2A} = 1$. $2(x-h)^2 - (y-2h)^2 = 2A$. This family has two parameters, $h$ and $A$. We need to differentiate twice. 2. First Differentiation: Differentiate with respect to $x$: $4(x-h) - 2(y-2h)y_1 = 0$, where $y_1=\frac{dy}{dx}$. $2(x-h) - (y-2h)y_1 = 0$. $2x-2h - yy_1 + 2hy_1 = 0$. $h(2y_1-2) = yy_1 - 2x \implies h = \frac{yy_1-2x}{2(y_1-1)}$. 3. Second Differentiation: Differentiate $2(x-h) - (y-2h)y_1 = 0$ with respect to $x$: $2 - [y_1 \cdot y_1 + (y-2h)y_2] = 0$, where $y_2=\frac{d^2y}{dx^2}$. $2 - y_1^2 - (y-2h)y_2 = 0$. 4. Eliminate h: From $2(x-h) = (y-2h)y_1$, we can express $(y-2h)$: $(y-2h) = \frac{2(x-h)}{y_1}$. Substitute this into the second derivative equation: $2 - y_1^2 - \frac{2(x-h)}{y_1} y_2 = 0$. Now we need to eliminate $h$. From $h = \frac{yy_1-2x}{2(y_1-1)}$, $x-h = x - \frac{yy_1-2x}{2(y_1-1)} = \frac{2x(y_1-1) - (yy_1-2x)}{2(y_1-1)} = \frac{2xy_1-2x-yy_1+2x}{2(y_1-1)} = \frac{y_1(2x-y)}{2(y_1-1)}$. Substitute this into the equation $2 - y_1^2 - \frac{2(x-h)}{y_1} y_2 = 0$: \[ 2 - y_1^2 - \frac{2}{y_1} \left( \frac{y_1(2x-y)}{2(y_1-1)} \right) y_2 = 0 \] \[ 2 - y_1^2 - \frac{2x-y}{y_1-1} y_2 = 0 \] \[ (2-y_1^2)(y_1-1) - (2x-y)y_2 = 0 \] \[ 2y_1 - 2 - y_1^3 + y_1^2 - (2x-y)y_2 = 0 \] \[ (y-2x)y_2 + y_1^2 + 2y_1 = y_1^3+2 \] This matches option (B). What if the transverse axis is vertical? Then $e^2 = 1+\frac{A}{B} \implies 3=1+\frac{A}{B} \implies \frac{A}{B}=2 \implies A=2B$. The equation is $\frac{(y-2h)^2}{B} - \frac{(x-h)^2}{2B} = 1 \implies 2(y-2h)^2-(x-h)^2=2B$. Differentiating: $4(y-2h)y_1 - 2(x-h)=0 \implies 2(y-2h)y_1=(x-h)$. Differentiating again: $2[y_1^2+(y-2h)y_2] - 1=0$. This leads to a different differential equation. However, the one derived from the horizontal transverse axis matches an option. Step 4: Final Answer
Assuming the hyperbolas have a horizontal transverse axis, the resulting differential equation is $(y-2x)y_2 + y_1^2 + 2y_1 = y_1^3+2$.