Question

If $(\alpha +1,\alpha )$ be a point interior to the regions of the parabola $y^2=4x$ bounded by the chord joining the points $(6,5)$ and $(7,4),$ then the total number of integral values of $\alpha$ is

Answer 1

For $(\alpha +1,\alpha )$ to lie inside the parabola,
${\alpha }^2-4(\alpha +1)<0$
$\Rightarrow \alpha \in (2-2\sqrt{2},2+2\sqrt{2})\cdots \left(1\right)$

Equation of chord passing through $(6,5)$ and $(7,4)$ is $x+y=11$
$(0,0)$ and $(\alpha +1,\alpha )$ should lie on the same side.
$0+0-11<0,$
$\therefore \alpha +1+\alpha -11<0$
$\Rightarrow \alpha <5\cdots \left(2\right)$



By $\left(1\right)$ and $\left(2\right),$
$\alpha \in (2-2\sqrt{2},2+2\sqrt{2})$
Integral values are : $0,1,2,3,4$