M, N are endpoints of a diameter 4x−y=15 of circle x^2 +y^2 −6x+6y−16=0 ; and are also on the tangents at the end points of the major axis of an ellipse respectively, such that MN is also tangent to the same ellipse at point P. If the major axis of the ellipse is along y=x, find eccentricity, length of latus rectum, centre and equation of derectrices.


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Question Number 28185 by ajfour last updated on 21/Jan/18

$M, N a r e e n d p o i n t s o f a d i a m e t e r$ $4 x - y = 15 o f c i r c l e$ $x^{2} + y^{2} - 6 x + 6 y - 16 = 0; a n d a r e$ $a l s o o n t h e t a n g e n t s a t t h e e n d$ $p o i n t s o f t h e m a j o r a x i s o f a n$ $e l l i p s e r e s p e c t i v e l y, s u c h t h a t$ $M N i s a l s o t a n g e n t t o t h e s a m e$ $e l l i p s e a t p o i n t P .$ $I f t h e m a j o r a x i s o f t h e e l l i p s e$ $i s a l o n g y = x, f i n d$ $e c c e n t r i c i t y, l e n g t h o f l a t u s$ $r e c t u m, c e n t r e a n d e q u a t i o n o f$ $d e r e c t r i c e s .$
	Answered by mrW2 last updated on 22/Jan/18

	Commented by mrW2 last updated on 22/Jan/18
	$circle: x^2 +y^2 −6x+6y−16=0 ⇔(x−3)^2 +(y+3)^2 =34 center C(3,−3), radius R=(√(34)) line: 4x−y=15 ⇔y=4x−15 4×3−(−3)=15 ⇒line passes through C. intersection of circle and line: (x−3)^2 +(4x−15+3)^2 =34 17(x−3)^2 =34 x−3=±(√2) ⇒x=3±(√2) ⇒y=12±4(√2)−15=−3±4(√2) M(3−(√2),−3−4(√2)), N(3+(√2),−3+4(√2)) distance from N to line y=x (i.e. x−y=0): d_N =((3+(√2)−(−3+4(√2)))/(√(1^2 +(−1)^2 )))=((6−3(√2))/(√2))=3((√2)−1) point G with MG⊥line y=x: x_G =x_M +λ y_G =y_M −λ ⇒y_M −λ=x_M +λ ⇒λ=((y_M −x_M )/2) ⇒x_G =x_M +((y_M −x_M )/2)=((x_M +y_M )/2)=((3−(√2)−3−4(√2))/2)=−((5(√2))/2) ⇒y_G =x_G =−((5(√2))/2) Eqn. of ellipse: the ellipse is following ellipse after a rotation of (π/4): (((x−k)^2 )/a^2 )+(y^2 /b^2 )=1 where b=d_N =3((√2)−1) Eqn. after rotation of (π/4): (((((x+y)/(√2))−k)^2 )/a^2 )+(((((−x+y)/(√2)))^2 )/b^2 )=1 (((x+y−(√2) k)^2 )/a^2 )+(((−x+y)^2 )/b^2 )=2 Point G is on the ellipse, (((−5(√2)−(√2) k)^2 )/a^2 )=2 (((5+k)^2 )/a^2 )=1 ⇒k=a−5 Line y=4x−15 is tangent of ellipse: (((x+4x−15−(√2) k)^2 )/a^2 )+(((−x+4x−15)^2 )/b^2 )=2 (((5x−15−(√2) k)^2 )/a^2 )+(((3x−15)^2 )/b^2 )=2 b^2 (5x−15−(√2)k)^2 +a^2 (3x−15)^2 =2a^2 b^2 b^2 [25x^2 −10(15+(√2)k)x+(15+(√2)k)^2 ]+a^2 [9x^2 −90x+225]=2a^2 b^2 (25b^2 +9a^2 )x^2 −10[(15+(√2)k)b^2 +9a^2 ]x+[(15+(√2)k)^2 b^2 +(225−2b^2 )a^2 ]=0 for a tangent there is only one intersection point, Δ=100[(15+(√2)k)b^2 +9a^2 ]^2 −4(25b^2 +9a^2 )[(15+(√2)k)^2 b^2 +(225−2b^2 )a^2 ]=0 k=a−5 b^2 =9(3−2(√2)) ⇒25[9(3−2(√2)){15+(√2)(a−5)}+9a^2 ]^2 −[225(3−2(√2))+9a^2 ][9(3−2(√2)){15+(√2) (a−5)}^2 +(171+36(√2))a^2 ]=0 ⇒((√2)−1)a^3 −10(3(√2) −4)a^2 =0 ⇒a=((10(3(√2)−4))/((√2)−1))=10(2−(√2))≈5.86 b=3((√2)−1)≈1.24 (b/a)=((3((√2)−1))/(10(2−(√2))))=((3(√2))/(20)) eccentricity=(c/a)=((√(a^2 −b^2 ))/a)=(√(1−((b/a))^2 ))=(√(1−(9/(200))))≈0.977 length of latus rectum =((2b^2 )/a)=((18(3−2(√2)))/(10(2−(√2))))=((18−9(√2))/(10)) k=a−5=5(3−2(√2))≈0.86 (√2)k=5(3(√2)−4) therefore the eqn. of ellipse (before rotation) is (([x−5(3−2(√2))]^2 )/(200(3−2(√2))))+(y^2 /(9(3−2(√2))))=1 center of ellipse is (k,0)=(5(3−2(√2)),0) the eqn. of the searched ellipse is (([x+y−5(3(√2)−4)]^2 )/(200))+(((y−x)^2 )/9)=2(3−2(√2)) center of ellipse is ((k/(√2)),(k/(√2)))=(((5(3(√2)−4))/2),((5(3(√2)−4))/2))$
	$c i r c l e :$ $x^{2} + y^{2} - 6 x + 6 y - 16 = 0$ $\Leftrightarrow {(x - 3)}^{2} + {(y + 3)}^{2} = 34$ $c e n t e r C (3, - 3), r a d i u s R = \sqrt{34}$ $l i n e :$ $4 x - y = 15$ $\Leftrightarrow y = 4 x - 15$ $4 \times 3 - (- 3) = 15$ $\Rightarrow l i n e p a s s e s t h r o u g h C .$ $i n t e r s e c t i o n o f c i r c l e a n d l i n e :$ ${(x - 3)}^{2} + {(4 x - 15 + 3)}^{2} = 34$ $17 {(x - 3)}^{2} = 34$ $x - 3 = \pm \sqrt{2}$ $\Rightarrow x = 3 \pm \sqrt{2}$ $\Rightarrow y = 12 \pm 4 \sqrt{2} - 15 = - 3 \pm 4 \sqrt{2}$ $M (3 - \sqrt{2}, - 3 - 4 \sqrt{2}), N (3 + \sqrt{2}, - 3 + 4 \sqrt{2})$ $d i s t a n c e f r o m N t o l i n e y = x (i . e . x - y = 0) :$ $d_{N} = \frac{3 + \sqrt{2} - (- 3 + 4 \sqrt{2})}{\sqrt{1^{2} + {(- 1)}^{2}}} = \frac{6 - 3 \sqrt{2}}{\sqrt{2}} = 3 (\sqrt{2} - 1)$ $p o i n t G w i t h M G ⊥ l i n e y = x :$ $x_{G} = x_{M} + λ$ $y_{G} = y_{M} - λ$ $\Rightarrow y_{M} - λ = x_{M} + λ$ $\Rightarrow λ = \frac{y_{M} - x_{M}}{2}$ $\Rightarrow x_{G} = x_{M} + \frac{y_{M} - x_{M}}{2} = \frac{x_{M} + y_{M}}{2} = \frac{3 - \sqrt{2} - 3 - 4 \sqrt{2}}{2} = - \frac{5 \sqrt{2}}{2}$ $\Rightarrow y_{G} = x_{G} = - \frac{5 \sqrt{2}}{2}$ $E q n . o f e l l i p s e :$ $t h e e l l i p s e i s f o l l o w i n g e l l i p s e a f t e r$ $a r o t a t i o n o f \frac{π}{4} :$ $\frac{{(x - k)}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $w h e r e b = d_{N} = 3 (\sqrt{2} - 1)$ $E q n . a f t e r r o t a t i o n o f \frac{π}{4} :$ $\frac{{(\frac{x + y}{\sqrt{2}} - k)}^{2}}{a^{2}} + \frac{{(\frac{- x + y}{\sqrt{2}})}^{2}}{b^{2}} = 1$ $\frac{{(x + y - \sqrt{2} k)}^{2}}{a^{2}} + \frac{{(- x + y)}^{2}}{b^{2}} = 2$ $P o i n t G i s o n t h e e l l i p s e,$ $\frac{{(- 5 \sqrt{2} - \sqrt{2} k)}^{2}}{a^{2}} = 2$ $\frac{{(5 + k)}^{2}}{a^{2}} = 1$ $\Rightarrow k = a - 5$ $L i n e y = 4 x - 15 i s t a n g e n t o f e l l i p s e :$ $\frac{{(x + 4 x - 15 - \sqrt{2} k)}^{2}}{a^{2}} + \frac{{(- x + 4 x - 15)}^{2}}{b^{2}} = 2$ $\frac{{(5 x - 15 - \sqrt{2} k)}^{2}}{a^{2}} + \frac{{(3 x - 15)}^{2}}{b^{2}} = 2$ $b^{2} {(5 x - 15 - \sqrt{2} k)}^{2} + a^{2} {(3 x - 15)}^{2} = 2 a^{2} b^{2}$ $b^{2} [25 x^{2} - 10 (15 + \sqrt{2} k) x + {(15 + \sqrt{2} k)}^{2}] + a^{2} [9 x^{2} - 90 x + 225] = 2 a^{2} b^{2}$ $(25 b^{2} + 9 a^{2}) x^{2} - 10 [(15 + \sqrt{2} k) b^{2} + 9 a^{2}] x + [{(15 + \sqrt{2} k)}^{2} b^{2} + (225 - 2 b^{2}) a^{2}] = 0$ $f o r a t a n g e n t t h e r e i s o n l y o n e i n t e r s e c t i o n p o i n t,$ $Δ = 100 {[(15 + \sqrt{2} k) b^{2} + 9 a^{2}]}^{2} - 4 (25 b^{2} + 9 a^{2}) [{(15 + \sqrt{2} k)}^{2} b^{2} + (225 - 2 b^{2}) a^{2}] = 0$ $k = a - 5$ $b^{2} = 9 (3 - 2 \sqrt{2})$ $\Rightarrow 25 {[9 (3 - 2 \sqrt{2}) {15 + \sqrt{2} (a - 5)} + 9 a^{2}]}^{2} - [225 (3 - 2 \sqrt{2}) + 9 a^{2}] [9 (3 - 2 \sqrt{2}) {15 + \sqrt{2} (a - 5)}^{2} + (171 + 36 \sqrt{2}) a^{2}] = 0$ $\Rightarrow (\sqrt{2} - 1) a^{3} - 10 (3 \sqrt{2} - 4) a^{2} = 0$ $\Rightarrow a = \frac{10 (3 \sqrt{2} - 4)}{\sqrt{2} - 1} = 10 (2 - \sqrt{2}) \approx 5.86$ $b = 3 (\sqrt{2} - 1) \approx 1.24$ $\frac{b}{a} = \frac{3 (\sqrt{2} - 1)}{10 (2 - \sqrt{2})} = \frac{3 \sqrt{2}}{20}$ $e c c e n t r i c i t y = \frac{c}{a} = \frac{\sqrt{a^{2} - b^{2}}}{a} = \sqrt{1 - {(\frac{b}{a})}^{2}} = \sqrt{1 - \frac{9}{200}} \approx 0.977$ $l e n g t h o f l a t u s r e c t u m = \frac{2 b^{2}}{a} = \frac{18 (3 - 2 \sqrt{2})}{10 (2 - \sqrt{2})} = \frac{18 - 9 \sqrt{2}}{10}$ $k = a - 5 = 5 (3 - 2 \sqrt{2}) \approx 0.86$ $\sqrt{2} k = 5 (3 \sqrt{2} - 4)$ $t h e r e f o r e t h e e q n . o f e l l i p s e (b e f o r e$ $r o t a t i o n) i s$ $\frac{{[x - 5 (3 - 2 \sqrt{2})]}^{2}}{200 (3 - 2 \sqrt{2})} + \frac{y^{2}}{9 (3 - 2 \sqrt{2})} = 1$ $c e n t e r o f e l l i p s e i s (k, 0) = (5 (3 - 2 \sqrt{2}), 0)$ $t h e e q n . o f t h e s e a r c h e d e l l i p s e i s$ $\frac{{[x + y - 5 (3 \sqrt{2} - 4)]}^{2}}{200} + \frac{{(y - x)}^{2}}{9} = 2 (3 - 2 \sqrt{2})$ $c e n t e r o f e l l i p s e i s (\frac{k}{\sqrt{2}}, \frac{k}{\sqrt{2}}) = (\frac{5 (3 \sqrt{2} - 4)}{2}, \frac{5 (3 \sqrt{2} - 4)}{2})$
	Commented by mrW2 last updated on 22/Jan/18

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