Find center, focii and intercepts on $x$ $x$ -axis and $y$ $y$ -axis of ellipse $16 x^{2} + 9 y^{2} + 64 x - 18 y - 71 = 0$ $16x^2+9y^2+64x-18y-71=0$ ?

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Find center, focii and intercepts on $x$ $x$ -axis and $y$ $y$ -axis of ellipse $16 x^{2} + 9 y^{2} + 64 x - 18 y - 71 = 0$ $16x^2+9y^2+64x-18y-71=0$ ?

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Answer 1 · 2017-07-07T08:59:26

Center is $(- 2, 1)$ $(-2,1)$ , focii are $(- 2, 1 - \sqrt{7})$ $(-2,1-sqrt7)$ and $(- 2, 1 + \sqrt{7})$ $(-2,1+sqrt7)$ . Intercepts on $x$ $x$ -axis are $1 \pm \frac{4 \sqrt{5}}{3}$ $1+-(4sqrt5)/3$ and intercepts on $y$ $y$ -axis are $- 2 \pm \frac{3 \sqrt{15}}{4}$ $-2+-(3sqrt15)/4$ .

Explanation:

The general form of equation of ellipse is

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2+(y-k)^2/b^2=1$ , whose center is $(h, k)$ $(h,k)$ and major axis is along $x$ $x$ -axis if $a > b$ $a>b$ and along $y$ $y$ -axis if $a < b$ $a< b$ .

Further eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$ $e=sqrt(1-b^2/a^2)$ , if $a > b$ $a>b$ or $e = \sqrt{1 - \frac{a^{2}}{b^{2}}}$ $e=sqrt(1-a^2/b^2)$ , if $a < b$ $a< b$ .

Focii are at major axis at a distance of $\pm a e$ $+-ae$ or $= \pm b e$ $=+-be$ , again depending on $a > b$ $a>b$ or $b > a$ $b> a$ .

We can express $16 x^{2} + 9 y^{2} + 64 x - 18 y - 71 = 0$ $16x^2+9y^2+64x-18y-71=0$ as

$16 x^{2} + 64 x + 9 y^{2} - 18 y - 71 = 0$ $16x^2+64x+9y^2-18y-71=0$

or $16 (x^{2} + 4 x + 4) + 9 (y^{2} - 2 y + 1) - 64 - 9 - 71 = 0$ $16(x^2+4x+4)+9(y^2-2y+1)-64-9-71=0$

or $16 {(x + 2)}^{2} + 9 {(y - 1)}^{2} = 144$ $16(x+2)^2+9(y-1)^2=144$

or $\frac{{(x + 2)}^{2}}{3^{2}} + \frac{{(y - 1)}^{2}}{4^{2}} = 1$ $(x+2)^2/3^2+(y-1)^2/4^2=1$

Hence center is $(- 2, 1)$ $(-2,1)$ and major axis along $y$ $y$ -axis is $8$ $8$ and minor axis along $x$ $x$ -axis is $6$ $6$ .

Eccentricity is $e = \sqrt{1 - {(\frac{3}{4})}^{2}} = \frac{\sqrt{7}}{4}$ $e=sqrt(1-(3/4)^2)=sqrt7/4$

and as $b e = \sqrt{7}$ $be=sqrt7$ , focii are $(- 2, 1 \pm \sqrt{7})$ $(-2,1+-sqrt7)$

Intercepts on $x$ $x$ -axis can be found by putting $y = 0$ $y=0$ i.e. $9 y^{2} - 18 y - 71 = 0$ $9y^2-18y-71=0$ $y = \frac{18 \pm \sqrt{324 - 4 \cdot 9 \cdot (- 71)}}{18} = 1 \pm \frac{\sqrt{324 + 2556}}{18} = 1 \pm 24 \frac{\sqrt{5}}{18} = 1 \pm \frac{4 \sqrt{5}}{3}$ $y=(18+-sqrt(324-4*9*(-71)))/18=1+-sqrt(324+2556)/18=1+-24sqrt5/18=1+-(4sqrt5)/3$

Intercepts on $y$ $y$ -axis can be found by putting $x = 0$ $x=0$ i.e. $16 x^{2} + 64 x - 71 = 0$ $16x^2+64x-71=0$ $x = \frac{- 64 \pm \sqrt{64^{2} - 4 \cdot 16 \cdot (- 71)}}{32} = - 2 \pm \frac{\sqrt{4096 + 4544}}{32} = - 2 \pm 24 \frac{\sqrt{15}}{32} = - 2 \pm \frac{3 \sqrt{15}}{4}$ $x=(-64+-sqrt(64^2-4*16*(-71)))/32=-2+-sqrt(4096+4544)/32=-2+-24sqrt15/32=-2+-(3sqrt15)/4$

graph{(16x^2+9y^2+64x-18y-71)((x+2)^2+(y-1-sqrt7)^2-0.03)((x+2)^2+(y-1+sqrt7)^2-0.03)=0 [-12.21, 7.79, -4.36, 5.64]}

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Find center, focii and intercepts on $x$ $x$ -axis and $y$ $y$ -axis of ellipse $16 x^{2} + 9 y^{2} + 64 x - 18 y - 71 = 0$ $16x^2+9y^2+64x-18y-71=0$ ?

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Explanation:

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Find center, focii and intercepts on $x$ $x$ -axis and $y$ $y$ -axis of ellipse $16 x^{2} + 9 y^{2} + 64 x - 18 y - 71 = 0$ $16x^2+9y^2+64x-18y-71=0$ ?