Question

Let the transverse axis of a hyperbola \( H \) be parallel to the X-axis and \( x^2 + y^2 - 2x - 4y + 3 = 0 \) be the equation of the auxiliary circle of \( H \). If the asymptotes of \( H \) are at right angles, then the equation of the hyperbola is:

• A. \(3x^2 - 2y^2 - 6x + 8y - 11 = 0\)
• B. \(x^2 - y^2 + 2x + 4y - 5 = 0\)
• C. \(3x^2 - 2y^2 + 6x + 8y - 11 = 0\)
• D. \(x^2 - y^2 - 2x + 4y - 5 = 0\)

Hint:

To check for rectangular hyperbola, verify that \(A + B = 0\) in the general conic form. Shift the origin to the center using completing square.

Answer 1

The Correct Option is D

The given auxiliary circle is: \[ x^2 + y^2 - 2x - 4y + 3 = 0 \] Completing the square: \[ (x - 1)^2 + (y - 2)^2 = 0^2 \Rightarrow \text{center is } (1,2), \text{ radius } = 0 \] But that leads to a point circle. Since that's not meaningful here, likely a mistake — let’s try again properly: \[ x^2 - 2x + y^2 - 4y + 3 = 0 \Rightarrow (x - 1)^2 + (y - 2)^2 = 0 \Rightarrow \text{auxiliary circle is a point circle} \] So the center of hyperbola = (1, 2), and we shift origin to this point. Let \( X = x - 1, \, Y = y - 2 \) For asymptotes to be at right angles, the general form must satisfy \(A + B = 0\) Try shifting and checking the options. Only option (4) satisfies the condition for rectangular hyperbola and center (1, 2).