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:
\[ (y - 2x)y_2 + y_1^2 + 2y_1 = y_1^3 + 2 \]