For ellipse 16x^2 +4y^2 +96x−8y−84=0 find i)centre ii)verteces iii)focus iv)directrix v)length of major and minor axis vi)ecentricity vii)graph the ellipse


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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

	$y e s s i r . . . p l e a s e c h e c k r i g h t$ $w r o n g ?$
	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

	$t h a n k s$
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