KVPY SX SB Exam 19-Nov-2017 Question Paper

Centre is $\frac{{\left(\mathit{x}-\frac{\mathit{\alpha }}{2}\right)}^2}{{\mathit{a}}^2}+\frac{{\left(\mathit{y}-\mathit{\beta }\right)}^2}{{\mathit{b}}^2}=1$
pass through (0, 0) (0, 2) & (1, 0)
Distance between
${\mathit{\alpha }}^2=4{\mathit{a}}^2{\mathit{e}}^2$
Pass through (0, 0)
⇒ $\frac{{\mathit{\alpha }}^2}{4{\mathit{a}}^2}+\frac{{\mathit{\beta }}^2}{{\mathit{b}}^2}=1$
$\frac{{\mathit{\alpha }}^2}{4{\mathit{a}}^2}={\mathit{e}}^2$
Pass through (0, 2)
$\frac{{\mathit{\alpha }}^2}{4{\mathit{a}}^2}+\frac{{\left(2-\mathit{\beta }\right)}^2}{{\mathit{b}}^2}=1$
from these two ⇒ $\mathit{\beta }=1$
Use properties $\mathit{P}{\mathit{F}}_1+\mathit{P}{\mathit{F}}_2=2\mathit{a}$
${\mathit{F}}_1\left(0,1\right),{\mathit{F}}_2\left(\mathit{\alpha },1\right),\mathit{P}\left(0,2\right)$
$1+\sqrt{\mathit{\alpha }+1}=2\mathit{a}$
${\mathit{F}}_1\left(0,1\right),{\mathit{F}}_2\left(\mathit{\alpha },1\right),\mathit{P}\left(1,0\right)$
$\sqrt{1+1}+\sqrt{{\left(\mathit{\alpha }-1\right)}^2+1}=2\mathit{a}$
From these two
$\mathit{\alpha }=-2\mathit{\alpha }+2\sqrt{2}\mathit{a}$
put α in any of above two equations
$\mathit{a}=\frac{\sqrt{2}+1}{2}\Rightarrow 2\mathit{a}=\sqrt{2}+1$
α = 1
α = 2ae
$\mathit{\alpha }=1,2\mathit{a}=\sqrt{2}+1$
find $\mathit{e}=\sqrt{2}$