Math  /  Algebra

QuestionWhich type of conic is represented by the equation below? 4x2+25y2+8x+150y+121=0-4 x^{2}+25 y^{2}+8 x+150 y+121=0
This is an equation of a hyperbola. Write the equation of this conic section in conic form.

Studdy Solution

STEP 1

1. The given equation is a conic section.
2. The equation is confirmed to represent a hyperbola.
3. The goal is to rewrite the equation in the standard form of a hyperbola.

STEP 2

1. Rearrange and group terms.
2. Complete the square for both xx and yy terms.
3. Rewrite the equation in standard form.

STEP 3

Rearrange and group the terms of the equation by variables:
4x2+8x+25y2+150y+121=0 -4x^2 + 8x + 25y^2 + 150y + 121 = 0
Group the xx terms and the yy terms:
(4x2+8x)+(25y2+150y)+121=0 (-4x^2 + 8x) + (25y^2 + 150y) + 121 = 0

STEP 4

Complete the square for the xx terms:
Factor out 4-4 from the xx terms:
4(x22x) -4(x^2 - 2x)
To complete the square, take half of the coefficient of xx, square it, and add and subtract it inside the parentheses:
4(x22x+11)=4((x1)21)=4(x1)2+4 -4(x^2 - 2x + 1 - 1) = -4((x - 1)^2 - 1) = -4(x - 1)^2 + 4

STEP 5

Complete the square for the yy terms:
Factor out 2525 from the yy terms:
25(y2+6y) 25(y^2 + 6y)
To complete the square, take half of the coefficient of yy, square it, and add and subtract it inside the parentheses:
25(y2+6y+99)=25((y+3)29)=25(y+3)2225 25(y^2 + 6y + 9 - 9) = 25((y + 3)^2 - 9) = 25(y + 3)^2 - 225

STEP 6

Substitute the completed squares back into the equation:
4(x1)2+4+25(y+3)2225+121=0 -4(x - 1)^2 + 4 + 25(y + 3)^2 - 225 + 121 = 0
Combine the constant terms:
4(x1)2+25(y+3)2100=0 -4(x - 1)^2 + 25(y + 3)^2 - 100 = 0

STEP 7

Rewrite the equation in standard form by isolating the constant:
4(x1)2+25(y+3)2=100 -4(x - 1)^2 + 25(y + 3)^2 = 100
Divide the entire equation by 100 to get the standard form of a hyperbola:
(y+3)24(x1)225=1 \frac{(y + 3)^2}{4} - \frac{(x - 1)^2}{25} = 1
This is the equation of the hyperbola in standard form.

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