Question

If the equation of the pair of lines passing through (1, 1) and perpendicular to the pair of lines \(2x^2 + xy - y^2 - x + 2y - 1 = 0\) is \(ax^2 + 2hxy + by^2 + 2gx + 3y = 0\). then \(\frac{b}{a} =\)

• A. \(g/h\)
• B. \(2(g+h)\)
• C. \(2(g-h)\)
• D. \(gh\)

Hint:

To find the equation of a pair of lines perpendicular to a given pair of lines \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\) and passing through a specific point \((x_0, y_0)\), use the transformation formula: \(B(x-x_0)^2 - 2H(x-x_0)(y-y_0) + A(y-y_0)^2 = 0\). Once you derive the transformed equation, compare its coefficients with the given form of the transformed equation to find the required values.

Answer 1

The Correct Option is B

Step 1: Understand the relationship between a pair of lines and a perpendicular pair of lines.
Let the equation of a pair of straight lines be \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\).
The equation of the pair of lines passing through a point \((x_0, y_0)\) and perpendicular to the given pair of lines is given by the formula: \[ B(x-x_0)^2 - 2H(x-x_0)(y-y_0) + A(y-y_0)^2 = 0. \] Step 2: Identify the coefficients from the given equation and the given point.
The given pair of lines is \(2x^2 + xy - y^2 - x + 2y - 1 = 0\). Comparing this to the general form \(Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0\), we have:
\(A = 2\)
\(2H = 1 \implies H = \frac{1}{2}\)
\(B = -1\)
The given point \((x_0, y_0)\) is \((1, 1)\).
Step 3: Substitute these values into the formula to find the equation of the perpendicular pair of lines. \[ -1(x-1)^2 - 2\left(\frac{1}{2}\right)(x-1)(y-1) + 2(y-1)^2 = 0 \] \[ -(x^2 - 2x + 1) - (xy - x - y + 1) + 2(y^2 - 2y + 1) = 0 \] Expand the products: \[ -x^2 + 2x - 1 - xy + x + y - 1 + 2y^2 - 4y + 2 = 0 \] Combine like terms: \[ -x^2 - xy + 2y^2 + (2x + x) + (y - 4y) + (-1 - 1 + 2) = 0 \] \[ -x^2 - xy + 2y^2 + 3x - 3y + 0 = 0 \] \[ -x^2 - xy + 2y^2 + 3x - 3y = 0. \] To match the standard form or to make the leading coefficient positive, we can multiply the entire equation by -1: \[ x^2 + xy - 2y^2 - 3x + 3y = 0. \] Step 4: Compare the derived equation with the given form of the transformed equation.
The problem states that the transformed equation is \(ax^2 + 2hxy + by^2 + 2gx + 3y = 0\).
Comparing \(x^2 + xy - 2y^2 - 3x + 3y = 0\) with \(ax^2 + 2hxy + by^2 + 2gx + 3y = 0\):
Coefficient of \(x^2\): \(a = 1\) Coefficient of \(xy\): \(2h = 1 \implies h = \frac{1}{2}\) Coefficient of \(y^2\): \(b = -2\) Coefficient of \(x\): \(2g = -3 \implies g = -\frac{3}{2}\) Coefficient of \(y\): \(3\) (This matches the given form) 
Step 5: Calculate the value of \(\frac{b}{a}\) and check the options.
From the identified coefficients, \(a=1\) and \(b=-2\).
So, \(\frac{b}{a} = \frac{-2}{1} = -2\).
Now, evaluate each option using the values of \(g = -\frac{3}{2}\) and \(h = \frac{1}{2}\): 
Option (1): \(\frac{g}{h} = \frac{-3/2}{1/2} = -3\).
Option (2): \(2(g+h) = 2\left(-\frac{3}{2} + \frac{1}{2}\right) = 2\left(-\frac{2}{2}\right) = 2(-1) = -2\). 
Option (3): \(2(g-h) = 2\left(-\frac{3}{2} - \frac{1}{2}\right) = 2\left(-\frac{4}{2}\right) = 2(-2) = -4\). 
Option (4): \(gh = \left(-\frac{3}{2}\right)\left(\frac{1}{2}\right) = -\frac{3}{4}\). 
The calculated value of \(\frac{b}{a} = -2\) matches the value of Option (2).