Question

: What are the coordinates of the vertices of the conic section shown below?

(y+2)^2 (x-3)^2
---------- - ---------- = 1
16 9

A: (0, -2) and (6, - 2)
B: (-2, -1) and (-2, 7)
C: (-1, -2) and (7, - 2)
D: (3, -6) and (3, 2)


: What are the coordinates of the foci of the conic section shown below?

(y+2)^2 (x-3)^2
---------- - ----------- = 1
25 4



a: (3 ± √21, -2)

b: (3 ± √29, -2)

c: (3, -2 ± √29)

d: (3, -2 ± √21)


: Which direction does the graph of the equation shown below open?

X^2 + 6x – 4y + 5 = 0

A: up
B: down
C: right
D: left

Answer 1

Answer on Question #57356 - Math - Analytic Geometry

Question 1. What are the coordinates of the vertices of the conic section shown below?

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

A: $(0; -2)$ and $(6; -2)$

B: $(-2; -1)$ and $(-2; 7)$

C: $(-1; -2)$ and $(7; -2)$

D: $(3; -6)$ and $(3; 2)$

Solution

This conic section is a hyperbola.

The equation of a hyperbola is:

$$\frac {(x - x _ {0}) ^ {2}}{a ^ {2}} - \frac {(y - y _ {0}) ^ {2}}{b ^ {2}} = \pm 1$$

So, we have "North-South opening hyperbola".

The coordinates of the center are $(3; -2)$. Then, the coordinates of the vertices of the hyperbola are $(3; -2 \pm b) \Leftrightarrow (3; -2 \pm 4) \Leftrightarrow (3; -6)$ and $(3; 2)$.

Answer: D. $(3; -6)$ and $(3; 2)$.

Question 2. What are the coordinates of the foci of the conic section shown below?

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

A: $(3 \pm \sqrt{21}; -2)$

B: $(3 \pm \sqrt{29}; -2)$

C: $(3; -2 \pm \sqrt{29})$

D: $(3; -2 \pm \sqrt{21})$

Solution

This conic section is a hyperbola.

The equation of a hyperbola is

$$\frac {(x - x _ {0}) ^ {2}}{a ^ {2}} - \frac {(y - y _ {0}) ^ {2}}{b ^ {2}} = \pm 1$$

So, we have "North-South opening hyperbola".

We have: $a = 2$ and $b = 5$. The coordinates of the center are $(3; -2)$. Therefore

$c = \sqrt{a^2 + b^2} = \sqrt{29}$. And the coordinates of the foci of the hyperbola are $(3; -2 \pm c) \Leftrightarrow (3; -2 \pm \sqrt{29})$.

Answer: $C(3; -2 \pm \sqrt{29})$.

Question 3. Which direction does the graph of the equation shown below open?

$$x ^ {2} + 6 x - 4 y + 5 = 0$$

A: up

B: down

C: right

D: left

Solution

Transform the expression:

$$x ^ {2} + 6 x - 4 y + 5 = x ^ {2} + 6 x + 9 - 9 - 4 y + 5 = (x + 3) ^ {2} - 4 y + 14 = 0$$

So, we have:

$$(x + 3) ^ {2} = 4 y - 14$$

The graph of the equation opens up.

Answer: A up.

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