If two lines $x+y=|a|$ and $ax=y+1$ intersect at a point which lies in fourth quadrant then find the minimum length of focal chord of the parabola $y^2=4a^2x+4|a-1|x+5$
Point of intersection in fourth quadrant gives me $a\in[0,1)$
So, parabola is $y^2=4(a^2-a+1)x+5$
I know that length of focal chord is $a(t+\frac1t)^2$ for $y^2=4ax$ with end end of focal chord being $(at^2,2at)$
Also, if the focal chord makes angle $\theta$ with x-axis then length of focal chord is $4a\csc^2\theta$
Don't know how to apply that here.
Best Answer
I think $a\in (-1,1)$.
Eliminating $y$ from $x+y=|a|$ and $ax=y+1$, we get $(a+1)x=|a|+1$. Since $a\not=-1$, we get $x=\frac{|a|+1}{a+1}$ and $y=\frac{a|a|-1}{a+1}$.
If $a\lt -1$, then $x=\frac{-a+1}{a+1}\lt 0$.
If $-1\lt a\lt 1$, then $x\gt 0$ and $y\leqslant \frac{a^2-1}{a+1}=a-1\lt 0$.
If $a\geqslant 1$, then $y=\frac{(a-1)(a+1)}{a+1}=a-1\geqslant 0$.
So, $a\in (-1,1)$ follows.
You can use the following :
The length of focal chord is $A(t+\frac 1t)^2$ for $y^2=4A(x-B)$ with end of focal chord being $(B+At^2,2At)$.
By AM-GM inequality, we have $$\begin{align}(a^2+|a-1|)\bigg(t+\frac 1t\bigg)^2&=(a^2+|a-1|)\bigg(t^2+\frac{1}{t^2}+2\bigg) \\\\&\geqslant (a^2+|a-1|)\bigg(2\sqrt{t^2\cdot\frac{1}{t^2}}+2\bigg) \\\\&=4a^2+4(1-a) \\\\&=4\bigg(a-\frac 12\bigg)^2+3 \\\\&\geqslant 3\end{align}$$
Therefore, the minimum length of focal chord is $\color{red}3$.