Question

If (-1,-1) is the point of intersection of the pair of lines \(2x^2+5xy-3y^2+2gx+2fy+c=0\) then g+f=

• A. 4c
• B. 3c
• C. 2c
• D. c

Hint:

The coordinates of the point of intersection of the pair of lines \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\) can be found by solving the simultaneous equations obtained by partial differentiation with respect to x and y. Also, the intersection point must satisfy the original equation itself.

Answer 1

The Correct Option is D

The equation is given as \(2x^2+5xy-3y^2+2gx+2fy+c=0\). This matches the standard form \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\), where \(a=2, 2h=5, b=-3\).
The point of intersection \( (x_0, y_0) = (-1, -1) \) must satisfy the equation.
Substituting the point into the equation:
\(2(-1)^2 + 5(-1)(-1) - 3(-1)^2 + 2g(-1) + 2f(-1) + c = 0 \).
\(2(1) + 5(1) - 3(1) - 2g - 2f + c = 0 \).
\(2 + 5 - 3 - 2g - 2f + c = 0 \).
\(4 - 2g - 2f + c = 0 \).
\(4 - 2(g+f) + c = 0 \).
The point of intersection also satisfies the partial derivative equations:
\(\frac{\partial F}{\partial x} = 4x + 5y + 2g = 0\). At (-1,-1): \(4(-1) + 5(-1) + 2g = 0 \implies -4-5+2g=0 \implies 2g=9 \implies g=9/2\).
\(\frac{\partial F}{\partial y} = 5x - 6y + 2f = 0\). At (-1,-1): \(5(-1) - 6(-1) + 2f = 0 \implies -5+6+2f=0 \implies 1+2f=0 \implies f=-1/2\).
Let's check the relation we found: \(4 - 2(g+f) + c = 0\).
\(g+f = 9/2 - 1/2 = 8/2 = 4\).
Substituting this into the relation: \(4 - 2(4) + c = 0 \implies 4 - 8 + c = 0 \implies c=4\).
We found \(g+f=4\) and \(c=4\). Therefore, \(g+f=c\).