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(D) The equation of the curve is $x^2 + hxy - y^2 + gx + fy = 0$. The equation of the line is $fx - gy = \lambda$,which implies $\frac{fx - gy}{\lambd

Vedclass · Accepted Answer

$\lambda$ may have any value

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(D) The equation of the curve is $x^2 + hxy - y^2 + gx + fy = 0$. The equation of the line is $fx - gy = \lambda$,which implies $\frac{fx - gy}{\lambda} = 1$. Making the equation of the curve homogeneous with the help of the line equation: $x^2 + hxy - y^2 + (gx + fy) \left( \frac{fx - gy}{\lambda} \right) = 0$. Multiplying by $\lambda$: $\lambda(x^2 + hxy - y^2) + (gx + fy)(fx - gy) = 0$. $\lambda x^2 + \lambda hxy - \lambda y^2 + gfx^2 - g^2xy + f^2xy - fgy^2 = 0$. $(\lambda + gf)x^2 + (\lambda h - g^2 + f^2)xy - (\lambda + fg)y^2 = 0$. For the lines to be mutually perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(\lambda + gf) + (-(\lambda + fg)) = 0$. $\lambda + gf - \lambda - fg = 0$. $0 = 0$. Since the condition is satisfied for any value of $\lambda$,$\lambda$ may have any value.

If the lines joining the origin to the points of intersection of the line $fx - gy = \lambda$ and the curve $x^2 + hxy - y^2 + gx + fy = 0$ are mutually perpendicular,then:

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