JEE Advance Conics

$Q(\cfrac { { t }^{ 2 } }{ 2 } ,2t)$ $FC:\quad y=\cfrac { 2t }{ { t }^{ 2 }-1 } (x-1)$ $QL:\quad L=\left| \cfrac { { t }^{ 3 } }{ \sqrt { 4{ t }^{ 2 }+{ ({ t }^{ 2 }-1) }^{ 2 } } } \right|$ $\left\{ \because { L }^{ 2 }=\left| \cfrac { 1 }{ \cfrac { 2 }{ { t }^{ 4 } } +\cfrac { 1 }{ { t }^{ 2 } } +\cfrac { 1 }{ { t }^{ 6 } } } \right| \right\}$ $\because \quad L\uparrow es\Rightarrow t\uparrow es$ $QL=max\{ at\quad t=3\ \ \ { L }_{ max }=\cfrac { 27 }{ 10 }$

Now using standard formula of focal length of chord ... which is essential for JEE aspect for quick calculation's , Hence :

${ L }_{ FC }=4a\csc ^{ 2 }{ \theta }$ where angle $\theta$ is angle made by FC with +ve x-axis ,

Note :: Here $a=1$ . and $P(9,6)$ & $S(1,0)$ hence

$\tan { \theta } =\cfrac { 3 }{ 4 }$

${ L }_{ FC }=\cfrac { 100 }{ 9 } \quad$

$\boxed { { QL }_{ max }+{ L }_{ FC }=\cfrac { 27 }{ 10 } +\cfrac { 100 }{ 9 } }$