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 $Q^2=4PR.$ 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 $Q^2=4PR.$  
 A: You are seeking the envelope of the family of lines. As described in the Wikipedia entry, there is a standard approach to this task, using Calculus.
Write $$F(\lambda) \;=\; \lambda^2 P + \lambda Q + R$$
so that differentiating with respect to $\lambda$ yields 
$$F^\prime(\lambda) \;=\; 2\lambda P + Q$$
Then simply eliminate $\lambda$ from the equations $F(\lambda) = 0$ and $F^\prime(\lambda) = 0$. Solving the second equation gives $\lambda = -\frac{Q}{2P}$; substituting into the first equation gives, after a little algebraic manipulation,
$$Q^2 = 4 PR$$
as desired. $\square$
A: This is a pretty brute-force method but it works.
Test
To figure out what was going on, I tried a simple case: $P=x$, $Q=y$, $R=1$.  The family of lines is $$\lambda^2x +\lambda y + 1 = 0$$ and the quadratic is $$y^2 = 4x$$  To say that the family of lines is tangent to the curve means that wherever they intersect, their derivatives agree.
So first, we look at the intersections.  If $\lambda^2 x + \lambda y + 1 = 0$, then $y = -\lambda x  -\frac{1}{\lambda}$ ($\lambda \neq 0$ because then the line's equation becomes $1=0$).  So $y^2 = \lambda^2 x^2 + 2x + \frac{1}{\lambda^2}$.  On the other hand, if $y^2 = 4x$, then
\begin{align*}
    \lambda^2 x^2 + 2x + \frac{1}{\lambda^2} &= 4x \\
\implies
    \lambda^2 x^2 - 2x + \frac{1}{\lambda^2} &= 0 \\
\implies \left(\lambda x - \frac{1}{\lambda}\right)^2 &= 0\\
\implies x &= \frac{1}{\lambda^2}
\end{align*}
If follows that $y = -\lambda \cdot \frac{1}{\lambda^2} - \frac{1}{\lambda} = -\frac{2}{\lambda}$.  This jibes with the second equation $y^2 = 4x$.
The derivative $\frac{dy}{dx}$ along the line $\lambda^2 x + \lambda y + 1 = 0$ is clearly the slope $-\lambda$.  The derivative along the curve $y^2=4x$ is
$$
    2y \frac{dy}{dx} = 4\implies \frac{dy}{dx} = \frac{2}{y}
$$
If $y=-\frac{2}{\lambda}$ then $\frac{2}{y} = -\lambda$, as desired.

The general case
Now I'm going to be careful not to divide by zero. Let $F = \lambda^2 P + \lambda Q + R$ and $G = Q^2-4PR$.  We want to show
$$
    \begin{vmatrix} F_x & F_y \\ G_x & G_y \end{vmatrix} =0
$$
at the intersections $F=G=0$.
First, we reduce the equations for the intersections.  If $\lambda^2 P + \lambda Q + R = 0$, then $\lambda Q = - \lambda^2 P - R$, so 
$$
\lambda^2 Q^2 = \lambda^4 P^2 + 2\lambda^2 PR + R^2
$$
On the other hand, if $Q^2 =4PR$, then $\lambda^2 Q^2 = 4\lambda PR$, and we have
\begin{align*}
    \lambda^4 P^2 + 2\lambda^2 PR + R^2 &= 4\lambda PR \\
\implies
    \lambda^4 P^2 - 2\lambda^2 PR + R^2 &= 0 \\
\implies
    \left(\lambda^2 P-R\right)^2 &=0 \\
\implies
    \lambda^2 P-R &=0 \tag{1}\\
\end{align*}
Substituting $R=\lambda ^2P$ into the first equation gives
$$
2\lambda^2 P + \lambda Q = 0 \implies \lambda(2\lambda P + Q) = 0 \implies 2\lambda P + Q=0 \tag{2}
$$
Now we compute the determinant of the derivatives:
$$
\begin{vmatrix} F_x & F_y \\ G_x & G_y \end{vmatrix}
= \begin{vmatrix}
      \lambda^2 P_x + \lambda Q_x + R_x &
      \lambda^2 P_y + \lambda Q_y + R_y \\
      2QQ_x - 4P_xR - 4PR_x &
      2QQ_y - 4P_yR - 4PR_y \\
  \end{vmatrix}
$$
Add $4P$ times the first line to the second line.
The determinant is
$$
  \begin{vmatrix}
      \lambda^2 P_x + \lambda Q_x + R_x &
      \lambda^2 P_y + \lambda Q_y + R_y \\
      2(2\lambda P+Q)Q_x +4(\lambda^2P- R)P_x  &
      2(2\lambda P+Q)Q_y +4(\lambda^2P- R)P_y 
  \end{vmatrix}
$$
In light of (1) and (2), the second line is zero.
A: 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.
