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Show that the lines joining the origin to the points of intersection of the line \mathrm{f x-g y=\lambda} and the curve \mathrm{x^2+h x y-y^2+g x+f y+c=0} are at right angle for all \mathrm{\lambda \in R} if c = 

Option: 1

\mathrm{c=0}


Option: 2

\mathrm{c=1}


Option: 3

\mathrm{c=-1}


Option: 4

\mathrm{c=\frac{1}{2}}


Answers (1)

best_answer

Line is \mathrm{\frac{f x-g y}{\lambda}=1}, Homogenizing we get

\begin{aligned} & \mathrm{x^2+h x y-y^2+(g x+f y) \quad \lambda+c \quad \lambda \quad=0} \\ \\& \Rightarrow \mathrm{\lambda^2\left(x^2+h x y-y^2\right)+\lambda(g x+f y)(f x-g y)+c(f x-g y)^2=0} \end{aligned}

Pair of line will be at right angle if coefficient of \mathrm{x^2+\text { coefficient of } y^2=0}

\mathrm{\Rightarrow\left(\lambda^2+\lambda g f+c f^2\right)+\left(-\lambda^2-\lambda f g+c g^2\right)=0 \Rightarrow c\left(f^2+g^2\right)=0} ,     when c = 0

Posted by

Ajit Kumar Dubey

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