Answer to Question #224028 in Analytic Geometry for Bless

Question

Reduce each equation to standard form. Then find the coordinates of the center, the foci, the ends of the major and minor axes, and the ends of each latus rectum. Sketch the curve

(a) x² + 4y² + 6x + 16y + 21 = 0

(b) 16x² + 4y² + 32x − 16y − 32 = 0

(c) 4x² + 8y² − 4x − 24y − 13 = 0

Answer 1

(a)

$x^2 + 4y^2 + 6x + 16y + 21 = 0$

$(x+3)^2+4(y+2)^2=4$

$\dfrac{(x+3)^2}{4}+\dfrac{(y+2)^2}{1}=1$

Center: $(-3, -2)$

$c=\sqrt{a^2-b^2}=\sqrt{4-1}=\sqrt{3}$

Foci: $(-3-\sqrt{3}, -2), (-3+\sqrt{3}, 2)$

Vertices: $(-5, -2), (-1, -2)$

Co-vertices: $(-3, -3), (-3, -1)$

First latus rectum: $x=-3-\sqrt{3}$

Second latus rectum: $x=-3+\sqrt{3}$

The length of the latera recta: $1$

$(-3-\sqrt{3}, -2.5), (-3-\sqrt{3}, -1.5)$

$(-3+\sqrt{3}, -2.5), (-3+\sqrt{3}, -1.5)$

(b)

$16x^2 + 4y^2 + 32x − 16y − 32 = 0$

$16x^2+32x+16+4y^2-16y+16=64$

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

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

Center: $(-1, 2)$

$c=\sqrt{a^2-b^2}=\sqrt{16-4}=2\sqrt{3}$

Foci: $(-1,2-2\sqrt{3}), (-1,2+2\sqrt{3})$

Vertices: $(-1, -2), (-1, 6)$

Co-vertices: $(-3,2), (1, 2)$

First latus rectum: $y=2-2\sqrt{3}$

Second latus rectum: $y=2+2\sqrt{3}$

The length of the latera recta: $1$

$(0, 2-2\sqrt{3}), (0,2+2\sqrt{3})$

$(1,2-2\sqrt{3}), (1, 2+2\sqrt{3})$

(c)

$4x^2 + 8y^2 − 4x − 24y − 13 = 0$

$4x^2-4x+1+8y^2-24y+18=32$

$4(x-\dfrac{1}{2})^2+8(y-\dfrac{3}{2})^2=32$

$\dfrac{(x-\dfrac{1}{2})^2}{8}+\dfrac{(y-\dfrac{3}{2})^2}{4}=1$

Center: $(\dfrac{1}{2},\dfrac{3}{2})$

$c=\sqrt{a^2-b^2}=\sqrt{8-4}=2$

Foci: $(-\dfrac{3}{2},\dfrac{3}{2}), (\dfrac{5}{2},\dfrac{3}{2})$

Vertices: $(\dfrac{1}{2}-2\sqrt{2},\dfrac{3}{2}),(\dfrac{1}{2}+2\sqrt{2},\dfrac{3}{2})$

Co-vertices: $(\dfrac{1}{2},-\dfrac{1}{2}), (\dfrac{1}{2},\dfrac{7}{2})$

First latus rectum: $x=-\dfrac{3}{2}$

Second latus rectum: $x=\dfrac{5}{2}$

The length of the latera recta: $2\sqrt{2}$

$(-\dfrac{3}{2}, \dfrac{3}{2}-2\sqrt{2}), (-\dfrac{3}{2},\dfrac{3}{2}+2\sqrt{2})$

$(\dfrac{5}{2},\dfrac{3}{2}-2\sqrt{2}), (\dfrac{5}{2}, \dfrac{3}{2}+2\sqrt{2})$