if the line px+qy=r tangents the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1, then 1) prove a^2 p^2 +b^2 q^2 =r^2 2) find the coordinates of the touching point.


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Question Number 157926 by mr W last updated on 29/Oct/21

$i f t h e l i n e p x + q y = r t a n g e n t s t h e$ $e l l i p s e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, t h e n$ $1) p r o v e a^{2} p^{2} + b^{2} q^{2} = r^{2}$ $2) f i n d t h e c o o r d i n a t e s o f t h e$ $t o u c h i n g p o i n t .$
	Answered by mindispower last updated on 29/Oct/21
	$.... paramatric of elips{(acos(t),bsin(t))),t∈[0,2π[ M′(t)=(−asin(t),bcos(t)) M(t)∈ Line pacos(t)+qbsin(t)=r...(1) M′(t) director vector and (−q,p) line director⇒ −qbcos(t)+pasin(t)=0... (1)^2 +(2)^2 ⇔p^2 a^2 cos^2 (t)+q^2 b^2 sin^2 (t)+2paqbsin(t)cos(t) +q^2 b^2 cos^2 (t)+p^2 a^2 sin^2 (t)−2abpqsin(t)cos(t)=r^2 +0 ⇔a^2 p^2 +b^2 q^2 =r^2 { ((−qbcos(t)+pasin(t)=0....∗sin(t)..1)),((pacos(t)+qbsin(t)=r......∗cos(t)....2)) :} (1) sin(t)cos(t)≠0 ⇒(1)+(2) pa=rcos(t)⇒cos(t)=((pa)/r) M=(((a^2 p)/r),((b^2 q)/r)) sin(t)=((qb)/r) sin(t)=0⇒−qbcos(t)=0,⇒q=0 pacos(t)=r⇒cos(t)=(r/(pa))⇒∣(r/(pa))∣=1 in this case the line is x=(r/p), vertical lign M((r/p),0) cos(t)=0⇒p=0 sin(t)=(r/(qb))......The lign is y=(r/q)=^ +_− 1 vertical line M(0,(r/q))$
	$. . . . p a r a m a t r i c o f e l i p s {(a c o s (t), b s i n (t))), t \in [0, 2 π [$ $M^{'} (t) = (- a s i n (t), b c o s (t))$ $M (t) \in L i n e p a c o s (t) + q b s i n (t) = r . . . (1)$ $M^{'} (t) d i r e c t o r v e c t o r a n d (- q, p) l i n e d i r e c t o r \Rightarrow$ $- q b c o s (t) + p a s i n (t) = 0 . . .$ ${(1)}^{2} + {(2)}^{2} \Leftrightarrow p^{2} a^{2} {c o s}^{2} (t) + q^{2} b^{2} {s i n}^{2} (t) + 2 p a q b s i n (t) c o s (t)$ $+ q^{2} b^{2} {c o s}^{2} (t) + p^{2} a^{2} {s i n}^{2} (t) - 2 a b p q s i n (t) c o s (t) = r^{2} + 0$ $\Leftrightarrow a^{2} p^{2} + b^{2} q^{2} = r^{2}$ ${\begin{cases} - q b c o s (t) + p a s i n (t) = 0 . . . . * s i n (t) . . 1 \\ p a c o s (t) + q b s i n (t) = r . . . . . . * c o s (t) . . . . 2 \end{cases}$ $(1) s i n (t) c o s (t) \neq 0$ $\Rightarrow (1) + (2) p a = r c o s (t) \Rightarrow c o s (t) = \frac{p a}{r}$ $M = (\frac{a^{2} p}{r}, \frac{b^{2} q}{r})$ $s i n (t) = \frac{q b}{r}$ $s i n (t) = 0 \Rightarrow - q b c o s (t) = 0, \Rightarrow q = 0$ $p a c o s (t) = r \Rightarrow c o s (t) = \frac{r}{p a} \Rightarrow∣ \frac{r}{p a} ∣= 1$ $i n t h i s c a s e t h e l i n e i s x = \frac{r}{p}, v e r t i c a l l i g n$ $M (\frac{r}{p}, 0)$ $c o s (t) = 0 \Rightarrow p = 0$ $s i n (t) = \frac{r}{q b} . . . . . . T h e l i g n i s y = \frac{r}{q} \overset{}{=} \underline{+} 1 v e r t i c a l l i n e$ $M (0, \frac{r}{q})$
	Commented by mr W last updated on 30/Oct/21

	$t h a n k s a l o t!$
	Commented by mindispower last updated on 02/Nov/21

	$p l e a s u r s i r h a v e a g o o d d a y$
	Answered by som(math1967) last updated on 30/Oct/21

	$p x + q y = r \Rightarrow y = \frac{r - p x}{q}$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $o r . \frac{x^{2}}{a^{2}} + \frac{{(r - p x)}^{2}}{b^{2} q^{2}} = 1$ $o r . x^{2} b^{2} q^{2} + a^{2} p^{2} x^{2} - 2 a^{2} p x r + r^{2} a^{2} = a^{2} b^{2} q^{2}$ $o r . (b^{2} q^{2} + a^{2} p^{2}) x^{2} - 2 a^{2} p x r + a^{2} r^{2} - a^{2} b^{2} q^{2} = 0 . . . 1)$ $p x + q y = r i s t a n g e n t s o f \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $∴ {(- 2 a^{2} p r)}^{2} - 4 (b^{2} q^{2} + a^{2} p^{2}) (a^{2} r^{2} - a^{2} b^{2} q^{2}) = 0$ $[f o r e q u a l r o o t s d i s c r i m i n a n t i s 0]$ $4 a^{2} [a^{2} p^{2} r^{2} - (b^{2} q^{2} + a^{2} p^{2}) (r^{2} - b^{2} q^{2})] = 0$ $a^{2} p^{2} r^{2} - b^{2} q^{2} r^{2} + b^{4} q^{4} - a^{2} p^{2} r^{2} + a^{2} b^{2} p^{2} q^{2} = 0$ $b^{2} q^{2} (b^{2} q^{2} + a^{2} p^{2} - r^{2}) = 0$ $∴ a^{2} p^{2} + b^{2} q^{2} = r^{2}$ $f r o m e q u n . 1$ $(b^{2} q^{2} + a^{2} p^{2}) x^{2} - 2 a^{2} p x r + a^{2} r^{2} - a^{2} b^{2} q^{2} = 0$ $[(a^{2} p^{2} + b^{2} q^{2} = r^{2}]$ $r^{2} x^{2} - 2 a^{2} p x r + a^{2} (r^{2} - b^{2} q^{2}) = 0$ $r^{2} x^{2} - 2 . r x . a^{2} p + a^{4} . p^{2} = 0$ $x = \frac{a^{2} p}{r}$ $y = \frac{r - p x}{q} = \frac{r - \frac{a^{2} p^{2}}{r}}{q} = \frac{r^{2} - a^{2} p^{2}}{q r}$ $(\frac{a^{2} p}{r}, \frac{r^{2} - a^{2} p^{2}}{q r})$ $(\frac{a^{2} p}{r}, \frac{b^{2} q}{r})$
	Commented by mr W last updated on 30/Oct/21

	$t h a n k s a l o t!$
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