Answer to Question #219520 in Analytic Geometry for Jessy

Question

Write the equation of the ellipse which satisfies the given conditions.

1a. Center (2, 3), horizontal axis 8, vertical axis 4

b. Center (1, -2), horizontal major axis 8, eccentricity

2a. Foci (-2,1) and (4, 1), major axis 10

b. Foci (-3,0) and (-3, 4), minor axis 6

3a. Foci (-2,2) and (4,2), eccentricity

b. Foci (+2, 0), directrices x = +8

4a. Focus (0,0), vertex (5,0), eccentricity 0.5

b. Focus (0,0), vertex (0, 2), eccentricity 0.6

5a. Center (1,3), V(1,-1), and passing through the origin

b. Center (1, 1), V(3, 1), and passing through the origin

Answer 1

1.

a)

$C(2, 3), 2a=8, 2b=4$

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

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

b)

$C(1, -2), 2a=8, \varepsilon=\dfrac{c}{a}=\dfrac{\sqrt{a^2-b^2}}{a}=\dfrac{\sqrt{3}}{2}$

$a^2-b^2=\dfrac{3}{4}a^2$

$b=\dfrac{a}{2}=2$

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

2.

a)

$F_1(-2,1), F_2(4, 1), 2a=10$

$a=5$

$k=1, h-c=-2, h+c=4$

$C(1, 1)$

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

$b=4$

$\dfrac{(x-1)^2}{25}+\dfrac{(y-1)^2}{16}=1$

b)

$F_1(-3,0), F_2(-3, 4), 2b=6$

$b=3$

$h=-3, k-c=0, k+c=4$

$C(-3, 2)$

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

$a=\sqrt{13}$

$\dfrac{(x+3)^2}{9}+\dfrac{(y-2)^2}{13}=1$

3.

a)

$F_1(-2,2), F_2(4, 2), \dfrac{c}{a}=\dfrac{3}{5}$

$k=2,h-c=-2, h+c=4$

$C(1, 2)$

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

$a=3(\dfrac{3}{5})=5, b=\sqrt{5^2-3^2}=4$

$\dfrac{(x-1)^2}{25}+\dfrac{(y-2)^2}{16}=1$

b)

$F(\pm2,0), x=\pm8$

$k=0,$

$h+ae=2, h-ae=-2$

$h+\dfrac{a}{e}=8, h-\dfrac{a}{e}=-8$

$h=0$

$a=8e, 8e^2=2$

$e=\dfrac{c}{a}=\dfrac{1}{2}, a=4, c=\sqrt{a^2-b^2}=2$

$C(0, 0)$

$b=2\sqrt{3}$

$\dfrac{x^2}{16}+\dfrac{y^2}{12}=1$

4.

a)

$F_1(0, 0)$

$V_1(5,0)$

$e=0.5$

$k=0,$

$h+ae=0, h+a=5$

$h+0.5a=0, h+a=5$

$a=10, h=-5$

or

$h-ae=0, h+a=5$

$h-0.5a=0, h+a=5$

$a=\dfrac{10}{3}, h=\dfrac{5}{3}$

$a=10, h=-5$

$C(-5,0)$

$c=\sqrt{a^2-b^2}=5$

$b=5\sqrt{3}$

$\dfrac{(x+5)^2}{100}+\dfrac{y^2}{75}=1$

$a=\dfrac{10}{3}, h=\dfrac{5}{3}$

$C(\dfrac{5}{3},0)$

$c=\sqrt{a^2-b^2}=\dfrac{5}{3}$

$b=\dfrac{5\sqrt{3}}{3}$

$\dfrac{(x-\dfrac{5}{3})^2}{100/9}+\dfrac{y^2}{75/9}=1$

b)

$F_1(0, 0)$

$V_1(0,2)$

$e=0.6$

$h=0,$ $k+ae=0, k+a=2$

$k+0.6a=0, k+a=2$

$a=5, k=-3$

or

$k-ae=0, k+a=2$

$k-0.6a=0, k+a=2$

$a=\dfrac{5}{4}, k=\dfrac{3}{4}$

$a=5, k=-3$

$C(0,-3)$

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

$b=4$

$\dfrac{x^2}{16}+\dfrac{(y+3)^2}{25}=1$

$a=\dfrac{5}{4}, k=\dfrac{3}{4}$

$C(0,\dfrac{3}{4})$

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

$b=1$

$\dfrac{x^2}{1}+\dfrac{(y-\dfrac{3}{4})^2}{25/16}=1$

5.

a)

Center $(1, 3)$

Vertices $(1, -1)$

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

Center $(h, k)$

Vertices $(h, k\pm a)$

$h=1, k=3$

$3-a=-1=>a=4$

Passing through the origin

$\dfrac{(0-1)^2}{b^2}+\dfrac{(0-3)^2}{4^2}=1$

$b^2=16/7$

$\dfrac{(x-1)^2}{16/7}+\dfrac{(y-3)^2}{16}=1$

b)

Center $(1, 1)$

Vertices $(3, 1)$

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

Center $(h, k)$

Vertices $(h\pm a, k)$

$h=1, k=1$

$1+a=3=>a=2$

Passing through the origin

$\dfrac{(0-1)^2}{2^2}+\dfrac{(0-1)^2}{b^2}=1$

$b^2=4/3$

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