TITLE: If a family of straight lines is $\lambda^2 P+\lambda Q+R=0$ ,then the family of lines will be tangent to the curve $Q^2=4PR.$
QUESTION [6 upvotes]: I have read this theorem in my book but i do not know how to prove it.
If a family of straight lines can be represented by an equation $\lambda^2 P+\lambda Q+R=0$ where $\lambda $ is a parameter and $P,Q,R$ are linear functions of $x$ and $y$ then the family of lines will be tangent to the curve $Q^2=4PR.$
My try:
Let $P:a_1x+b_1y+c_1=0$
$Q:a_2x+b_2y+c_2=0$
$R:a_3x+b_3y+c_3=0$
Then the family of straight lines can be represented by an equation $\lambda^2 (a_1x+b_1y+c_1)+\lambda (a_2x+b_2y+c_2)+(a_3x+b_3y+c_3)=0$
$x(\lambda^2a_1+\lambda a_2+a_3)+y(\lambda^2b_1+\lambda b_2+b_3)+(\lambda^2c_1+\lambda c_2+c_3)=0$
But i do not know how to prove that the family of lines will be tangent to the curve $Q^2=4PR.$

REPLY [2 votes]: To find envelope by C-discriminant method, we eliminate $\lambda$ between
$$f(\lambda) \;=\; \lambda^2 P + \lambda Q + R =0, \; f^\prime(\lambda) \;=\; 2\lambda P + Q =0 \; $$
( the latter is the characteristic for the parameter $\lambda$ )
yielding:
$$ Q^2 = 4 P R $$
The method can be can be extended to 2,3 .. arbitrary number of parameters.
The method and result are same as that given by Blue.