JEE Advanced 2020 Paper 2

Since Normal at point P makes equal intercept on co-ordinate axes, therefore slope of Normal $=1$ Hence slope of tangent $y-0=1(x-1)$ Equation of tangent $y=x-1$ $(x_{1} y_{1})$ Equation of tangent at $\;\frac{x x_{1}}{a^{2}}-\;\frac{y y_{1}}{b^{2}}=1$ $x - y =1$ $x_{1}=a^{2}, y_{1}- b^{2}$ (equation of Tangent) on comparing $a^{2}-b^{2}=1 \;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot(1)$ Also $(x_{1} y_{1})=(a_{1}^{2} b^{2})$ Now equation of normal at $y-b^{2}=-1(x-a^{2})$ $x + y = a^{2}+ b^{2} \dots$ $x$ (Normal) point of intersection with $(a^{2}+b^{2})$ -axis is $e = \sqrt{1+\;\frac{b^{2}}{a^{2}}}$ Now $e=\sqrt{1+\;\frac{b^{2}}{b^{2}+1}}[$ $(1)\;\frac{b^{2}}{b^{2}+1}<1]$ from $1 < e < \sqrt{2}$ $\Delta =\;\frac{1}{2}\cdot AB\cdot PQ$ option (A) $\Delta =\;\frac{1}{2}\cdot AB\cdot PQ$ $\;\text{and}\;\Delta =\;\frac{1}{2}(a^{2}+b^{2}-1)\cdot b^{2}$ $\Delta =\;\frac{1}{2}(2 b^{2}) b^{2} \;\text{(from}\;\text{(1)}\;a^{2}-1=b^{2}\;\text{)}\;$ $\Delta = b^{4}$ $1 < e < \sqrt{2}$ so option ( D )