Answer to Question #221983 in Analytic Geometry for Celia

Question

16x²+4y²+32x-16y-32=0

center,foci, ends of major and minor axis, ends of latus rectum

Answer 1

Consider the equation $16x^2+4y^2+32x-16y-32=0$

$16(x^2+2x)+4(y^2-4y)=32$

$16(x^2+2x+1-1)+4(y^2-4y+4-4)=32$

$16(x^2+2x+1)-16+4(y^2-4y+4)-16=32$

$16(x+1)^2+4(y-2)^2=32+16+16$

$16(x+1)^2+4(y-2)^2=64$

Dividing by 64, we get

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

Observe that the above equation is of the form

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

So, the given equation is ellipse

Center of the ellipse $=(h,k)=(-1,2)$

And $a^2=4,b^2=16$

Observe that $a^2<b^2$

To find the foci of the ellipse, find $c$ using the relation $c^2=b^2-a^2$

$c^2=16-4=12$

$\Rightarrow c=\pm 2\sqrt{3}$

Foci of the ellipse are: $(h,k-\sqrt{c}),(h,k+\sqrt{c})=(-1,2-2\sqrt{3}),(-1,2+2\sqrt{3})$

Ends of the major axis are: $(h,k-b),(h,k+b)=(-1,2-4),(-1,2+4)=(-1,-2),(-1,6)$

Ends of the minor axis are:

$(h-a,k),(h+a,k)=(-1-2,2),(-1+2,2)=(-3,2),(1,2)$

Ends of the latus rectum are:

$(\frac{-a^2}{b},c),(\frac{a^2}{b},c)=(-1,2\sqrt{3}),(1,2\sqrt{3})$