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Question Number 30377 by ajfour last updated on 21/Feb/18

Commented by ajfour last updated on 21/Feb/18

$G i v e n t h e e l l i p s e t o u c h e s p a r a b o l a$ $o n l y a t v e r t e x a n d l i e s w i t h i n t h e$ $t h e r e d l i n e a n d p a r a b o l a .$
	Answered by mrW2 last updated on 22/Feb/18

	$E q n . o f e l l i p s e :$ ${(\frac{x}{a})}^{2} + {(\frac{2 y}{x_{0}} - 1)}^{2} = 1$ $I n t e r s e c t i o n w i t h p a r a b o l a :$ $\frac{x_{0} y}{a^{2}} = 1 - {(\frac{2 y}{x_{0}} - 1)}^{2} = \frac{4 y}{x_{0}} (1 - \frac{y}{x_{0}})$ $y [\frac{x_{0}}{a^{2}} - \frac{4}{x_{0}} + \frac{4 y}{x_{0}^{2}}] = 0$ $\Rightarrow y = 0$ $\Rightarrow \frac{x_{0}}{a^{2}} - \frac{4}{x_{0}} + \frac{4 y}{x_{0}^{2}} = 0$ $\Rightarrow y = x_{0} - \frac{x_{0}^{3}}{4 a^{2}} = x_{0} (1 - \frac{x_{0}^{2}}{4 a^{2}}) \overset{!}{=} 0$ $\Rightarrow {(2 a)}^{2} = x_{0}^{2}$ $\Rightarrow a = \frac{x_{0}}{2}$ $i . e . t h e l a r g e s t e l l i p s e i s a c i r c l e :$ ${(\frac{2 x}{x_{0}})}^{2} + {(\frac{2 y}{x_{0}} - 1)}^{2} = 1$ $w i t h r a d i u s \frac{x_{0}}{2} a n d i t s a r e a i s \frac{π x_{0}^{2}}{4} .$
	Commented by ajfour last updated on 21/Feb/18

	$G o o d w a y S i r .$
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