Question

Consider the parabola $x=ay-b{y}^2$ (where $b\ne 0$) If the exhaustive set of values of a for which there exist $\alpha ,\beta ϵR-\left\{0\right\}$ such that both the point $(\alpha ,\beta )$ and $(\beta ,\alpha )$ lies on the given parabola is $(-\infty ,p)\cup (q,\infty )$ then $\frac{p^2+q^2}{4}$ is

Answer 1

$(\alpha ,\beta )$ and $(\beta ,\alpha )$ will lies an some line $y=-x+\lambda$

solving line and parabola
$x=a(\lambda -x)-b(\lambda -x{)}^2$

$\Rightarrow b{x}^2+(a-2b\lambda +1)x+b{\lambda }^2-a\lambda =0\dots \left(1\right)$

Putting value of x is parabola

$\lambda -y=ay-b{y}^2$

$b{y}^2-y(a+1)+\lambda =0\dots \left(2\right)$

equation (1) & (2) have same roots $\alpha ,\beta$ so equation are identical.

So $\frac{b}{b}=\frac{a-2b\lambda +1}{-(a+1)}=\frac{b{\lambda }^2-a\lambda }{\lambda }$

so, $b\lambda -a=1\Rightarrow \lambda =\frac{1+a}{b}\dots \left(3\right)$

equation (2) has real roots so D>0

$(a+1{)}^2-4b\lambda >0$
$a^2+2a+1-4(1+a)>0$ (using equation (3))

$\Rightarrow (a-3)(a+1)>0$

$\Rightarrow aϵ(-\infty ,-1)\cup (3,\infty )$

so $p=-1,q=3$

$\frac{p^2+q^2}{4}=\frac{10}{4}=2.50$