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: Formulate the Equation of the Hyperbola Family Let the centre of the hyperbola be \( (h, k) \). Since the centre lies on the line \( y = 2x \), we have \( k = 2h \). The axes are parallel to the coordinate axes. The standard equation is: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] Given eccentricity \( e = \sqrt{3} \). Using \( e^2 = 1 + \frac{b^2}{a^2} \): \[ (\sqrt{3})^2 = 1 + \frac{b^2}{a^2} \implies 3 = 1 + \frac{b^2}{a^2} \implies b^2 = 2a^2 \] Substituting \( k=2h \) and \( b^2=2a^2 \) into the equation: \[ \frac{(x-h)^2}{a^2} - \frac{(y-2h)^2}{2a^2} = 1 \] Multiplying by \( 2a^2 \): \[ 2(x-h)^2 - (y-2h)^2 = 2a^2 \] Let \( 2a^2 = C \) (an arbitrary constant). \[ 2(x-h)^2 - (y-2h)^2 = C \]
Step 2: Differentiate to Eliminate Constants Differentiating with respect to \( x \): \[ 4(x-h) - 2(y-2h)y_1 = 0 \] \[ 2(x-h) = (y-2h)y_1 \quad \dots(1) \] Differentiating again with respect to \( x \): \[ 2(1) - [ y_1 \cdot y_1 + (y-2h)y_2 ] = 0 \] \[ 2 - y_1^2 - (y-2h)y_2 = 0 \] \[ (y-2h) = \frac{2 - y_1^2}{y_2} \quad \dots(2) \]
Step 3: Eliminate Parameter \( h \) Substitute \( (y-2h) \) from eq (2) into eq (1): \[ 2(x-h) = \left( \frac{2 - y_1^2}{y_2} \right) y_1 \] \[ x-h = \frac{2y_1 - y_1^3}{2y_2} \quad \dots(3) \] We need to form a relation involving \( y-2x \). Note that: \[ y - 2x = (y-2h) - 2(x-h) \] Substitute expressions from (2) and (3): \[ y - 2x = \frac{2 - y_1^2}{y_2} - 2\left( \frac{2y_1 - y_1^3}{2y_2} \right) \] \[ y - 2x = \frac{2 - y_1^2}{y_2} - \frac{2y_1 - y_1^3}{y_2} \] \[ y - 2x = \frac{2 - y_1^2 - 2y_1 + y_1^3}{y_2} \] Multiplying by \( y_2 \): \[ (y - 2x)y_2 = y_1^3 - y_1^2 - 2y_1 + 2 \] Rearranging terms to match options: \[ (y - 2x)y_2 + y_1^2 + 2y_1 = y_1^3 + 2 \]