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Question Number 51492 by peter frank last updated on 27/Dec/18

F o r e l l i p s e

16 x^{2} + 4 y^{2} + 96 x - 8 y - 84 = 0

f i n d

i) c e n t r e

i i) v e r t e c e s

i i i) f o c u s

i v) d i r e c t r i x

v) l e n g t h o f m a j o r

a n d m i n o r a x i s

v i) e c e n t r i c i t y

v i i) g r a p h t h e e l l i p s e

Answered by peter frank last updated on 27/Dec/18

16 x^{2} + 96 x + 4 y^{2} - 8 y - 84 = 0

16 (x^{2} + 6 x + 9) + 4 (y^{2} - 2 y + 1) = 84 + 144 + 4

16 {(x + 3)}^{2} + 4 {(y - 1)}^{2} = 232

\frac{{(x + 3)}^{2}}{\frac{232}{16}} + \frac{{(y - 1)}^{2}}{\frac{232}{4}} = 1

a = \frac{\sqrt{232}}{4} b = \frac{\sqrt{232}}{2}

a^{2} = b^{2} (1 - e^{2})

e = \frac{3}{4}

\frac{(x - p)}{a^{2}} + \frac{(y - q)}{b^{2}} = 0

c e n t r e = (p, q) = (- 3, 1)

v e r t e c e s = (\pm a, 0) + (p, q)

v e r t e c e s = \pm (0.808, 1)

f o c u s = \pm (a e, 0) + (p, q)

f o c u s = \pm (0.264, 1)

d e r e c t r i x

x - p = \pm \frac{a}{e} = 3 \pm \frac{\sqrt{232}}{3}

l e n g t h o f m a j o r = 2 a = \frac{\sqrt{232}}{2} u n i t

m i n o r a x i s = 2 b = \sqrt{232} u n i t

Commented by ajfour last updated on 27/Dec/18

g r e a t, u^{'} v e s o l v e d y o u r s e l v e s!

Commented by peter frank last updated on 28/Dec/18

y e s s i r \dots p l e a s e c h e c k r i g h t

w r o n g ?

Commented by peter frank last updated on 28/Dec/18

m r W a n d a j f o u r h e l p t o

g r a p h t h e e l l i p s e

Commented by mr W last updated on 28/Dec/18

c o r r e c t s i r!

Commented by peter frank last updated on 28/Dec/18

t h a n k s