MATHS 3 Marks Question

$\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$

$\Rightarrow2\text{b}^2=\text{a}^2$

$\Rightarrow\text{a}^2=2\text{b}^2$

Now, $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)$

$\Rightarrow{\text{b}^2}=2\text{b}^2\big(1-\text{e}^2\big)$ $\big[\because\text{a}^2=2{\text{b}^2}\big]$

$\Rightarrow1=2\big(1-\text{e}^2\big)$

$\Rightarrow1=2-2\text{e}^2$

$\Rightarrow2\text{e}^2=2-1$

$\Rightarrow\text{e}^2=\frac{1}{2}$

$\Rightarrow\text{e}=\frac{1}{\sqrt{2}}$

$\therefore\text{ SP}=\frac{4}{5}\text{PM}$

$\Rightarrow\text{SP}^2=\frac{16}{25}(\text{PM})^2$

$\Rightarrow25\text{ SP}^2=16\text{ PM}^2$

$\Rightarrow25\bigg[\big(\text{x}+2\big)^2+\big(\text{y}-3\big)^2\bigg]=16\bigg[\frac{2\text{x}+3\text{y}+4}{\sqrt{2^2+3^2}}\bigg]^2$

$\Rightarrow25\big[\text{x}^2+4+4\text{x}+\text{y}^2+9-6\text{y}\big]=\frac{16\big(2\text{x}+3\text{y}+4\big)^2}{13}$

$\Rightarrow325\big[\text{x}^2+\text{y}^2+4\text{x}-6\text{y}+13\big]=16\big(2\text{x}+3\text{y}+4\big)^2$

$\frac{(\text{x}+2)^2}{\text{a}^2}+\frac{(\text{y}-3)^2}{\text{b}^2}=1$ $\big[\because\ $centre: (-2, 3) $\dots(\text{i})\ \big]$

$\Rightarrow\text{a}^2=4$

$\Rightarrow\text{b}^2=4$

$\frac{(\text{x}+2)^2}{4}+\frac{(\text{y}-3)^2}{9}=1$

$\Rightarrow\frac{9(\text{x}+2)^2+4(\text{y}-3)^2}{36}=1$

$\Rightarrow9(\text{x}+2)^2+4(\text{y}-3)^2=36$

$\Rightarrow9\big[\text{x}^2+4+4\text{x}\big]+4\big[\text{y}^2+9-6\text{y}\big]=36$

$\Rightarrow9\text{x}^2+36+36\text{x}+4\text{y}^2+36-24\text{y}=36$

$\Rightarrow9\text{x}^2+4\text{y}^2+36\text{x}-24\text{y}+36+36-36=0$

$\Rightarrow9\text{x}^2+4\text{y}^2+36\text{x}-24\text{y}+36=0$

$\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$

$\Rightarrow2\text{b}^2=\text{ab}$

$\Rightarrow\frac{\text{b}^2}{b}=\frac{\text{a}}{2}$

$\Rightarrow\text{b}=\frac{\text{a}}{2}$

$\Rightarrow\text{b}^2=\frac{\text{a}^2}{4}$

Now, $\text{b}^2=\text{a}^2\big(1-\text{e}^2\big)$

$\Rightarrow\frac{\text{a}^2}{4}=\text{a}^2\big(1-\text{e}^2\big)$ $\Big[\because\text{b}^2=\frac{\text{a}^2}{4}\Big]$