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

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

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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