For the conic section the equation

Question

25x²+50x+169y²-4200=0

what is true 

a.the lenght of the conjugate axis is 10

b. The graph has a vertex at (12,0)

c. The graph has a focus at (13,0

d the center of the graph is at (0,1)

Answer 1

There a squared terms for both x and y, and they both have the same sign.

 

That means that the curve is a circle or an ellipse.

 

Also, the coefficients of x² and y² are different.  That means that it is an ellipse.

 

That eliminates a, because of all the conic sections, only a hyperbola has a conjugate axis.

 

However, any of the others are possible.  That means that we have to figure out the actual graph.

 

First let's complete the square for the x terms:

 

25x² + 50x + 169y² - 4200 = 0

25 (x² + 2x) + 169y² = 4200

25 (x² + 2x + 1) + 169y² = 4200 + 25

25 (x + 1)² + 169y² = 4225

 

Divide both sides by 4225:

(25 / 4225) * (x + 1)² + (169/4225) y² = 1

(1/169) * (x + 1)² + (1/25) y² = 1

(x + 1)² / 13² + y² / 5² = 1

 

So the curve is an ellipse centered at (-1, 0), with a horizontal major axis of 26 and a vertical minor axis of 10.

 

And the vertices of the ellipse are at (-1 + 13, 0) and (-1 -13, 0), or (12, 0) and (-14, 0).

 

One of these vertices corresponds to answer b.