Question

find the center,foci,vertices, endpoints of conjugate axis.determine the equation of the asymptotes and sketsch the graph

1. (y+6)²/25-(x-4)²/39=1

2. 9x²+126x-16y²-96y+153=0

Answer 1

1.

$$\dfrac{(y+6)^2}{25}-\dfrac{(x-4)^2}{39}=1$$

Center: $(4, -6)$

Vertices: $(4, -11), (4, -1)$

$$c^2=a^2+b^2=25+39=64=>c=8$$

Foci: $(4,-14), (4, 2)$

Endpoints of conjugate axis: $(4-\sqrt{39}, -6), (4+\sqrt{39}, -6)$

First asymptote: $y=-\dfrac{5\sqrt{39}}{39}(x-4)-6$

Second asymptote: $y=\dfrac{5\sqrt{39}}{39}(x-4)-6$

2.

$$9x^2+126x-16y^2-96y+153=0$$

$$9(x^2+14x+49)-16(y^2+6y+9)=144$$

$$\dfrac{(x+7)^2}{16}-\dfrac{(y+3)^2}{9}=1$$

Center: $(-7, -3)$

Vertices: $(-11, -3), (-3, -3)$

$$c^2=a^2+b^2=16+9=25=>c=5$$

Foci: $(-12,-3), (-2,-3)$

Endpoints of conjugate axis: $(-7, -6), (-7,0)$

First asymptote: $y=-\dfrac{3}{4}x-\dfrac{33}{4}$

Second asymptote: $y=\dfrac{3}{4}x+\dfrac{9}{4}$