SOLUTION: Determine the center, vertices, and foci for the following ellipse 18x^2+4y^2-108x+16y=106 Write word or phrase that best completes each statement or answers question

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Determine the center, vertices, and foci for the following ellipse 18x^2+4y^2-108x+16y=106 Write word or phrase that best completes each statement or answers question      Log On


   

Question 746555: Determine the center, vertices, and foci for the following ellipse 18x^2+4y^2-108x+16y=106
Write word or phrase that best completes each statement or answers question
Found 2 solutions by MathLover1, KMST:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
18x%5E2%2B4y%5E2-108x%2B16y=106+
%2818x%5E2-108x%2B__%29%2B%284y%5E2%2B16y%2B___%29-106+=0....complete the square
%2818x%5E2-108x%2B162%29-162%2B%284y%5E2%2B32y%2B16%29-16-106=0
%2818x%5E2-108x%2B162%29%2B%284y%5E2%2B32y%2B16%29-284=0.......factor out 18 from first group and 4 from second group
18%28x%5E2-6x%2B9%29%2B4%28y%5E2%2B8y%2B4%29-284=0

18%28x-3%29%5E2%2B4%28y%2B2%29%5E2=284

18%28x-3%29%5E2%2F284%2B4%28y%2B2%29%5E2%2F284=284%2F284

%28x-3%29%5E2%2F15.78%2B%28y%2B2%29%5E2%2F71=1...round 15.78 to whole number

%28x-3%29%5E2%2F16%2B%28y%2B2%29%5E2%2F71=1....=> h=3 and k=-2

the center is at: (3,-2)
vertices: since the y term has the larger denominator, the ellipse is vertical and
vertices (3, -2%2B-sqrt%2871%29 ≅ (3, -6.43) and (3,-+10.43)
and foci:
(3, -2%2B-sqrt%28497%29%2F3 ≅ (3, -2%2B-7.43 ≅
(3, -5.43) and (3,-+9.43)




Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
An equation of the form
%28x-h%29%5E2%2F%28a%5E2%29%2B%28y-k%29%5E2%2F%28b%5E2%29=1 represents an ellipse centered at (h,k), with an axis of length 2a parallel to the x-axis,
and an axis of length 2b parallel to the y-axis.
If we can transform the equation given into such a form, we will be able to find everything the problem asks for.

18x%5E2%2B4y%5E2-108x%2B16y=106 --> 18x%5E2-108x%2B4y%5E2%2B16y=106 --> %2818x%5E2-108x%29%2B%284y%5E2%2B16y%29=106 --> 18%28x%5E2-6x%29%2B4%28y%5E2%2B4y%29=106
At this point, you look at the two expressions in brackets and have to realize that we can add something to each expression to "complete a square"
x%5E2-6x is part of x%5E2-6x%2B9=%28x-3%29%5E2 and
y%5E2%2B4y is part of y%5E2%2B4y%2B4=%28y%2B2%29%5E2
So 18%28x-3%29%5E2=18%28x%5E2-6x%2B9%29=18x%5E2-108x%2Bhighlight%28162%29 and
4%28y%2B2%29%5E2=4%28y%5E2%2B4y%2B4%29=4y%5E2%2B16y%2Bhighlight%2816%29
So we go back to the original equation, and add 162%2B16 to both sides of the equal sign:
18x%5E2%2B4y%5E2-108x%2B16y=106 --> 18x%5E2-108x%2B162%2B4y%5E2%2B16y%2B16=106%2B162%2B16 --> %2818x%5E2-108x%2B162%29%2B%284y%5E2%2B16y%2B16%29=284 --> 18%28x%5E2-6x%2B9%29%2B4%28y%5E2%2B4y%2B4%29=284 --> 18%28x-3%29%5E2%2B4%28y%2B2%29%5E2=284
Dividing both sides of the equal sign by 284 the equation turns into
%28x-3%29%5E2%2F%28%28284%2F18%29%29%2B%28y%2B2%29%5E2%2F%28%28284%2F4%29%29=1

That is the equation of an ellipse with highlight%28center%29 at (3,-2).
The axis parallel to the y-axis (along x=3) is longer, and it is called the major axis.
Half of its length (called the semi-major axis) is
sqrt%28284%2F4%29=sqrt%2871%29
The %28vertices%29 are the ends of the major axis, at a distance sqrt%2871%29 from the center, and are at
(3,-2-sqrt%2871%29) and (3,-2%2Bsqrt%2871%29)

The other axis is called the minor axis.
It is along the line y=-2, parallel to the x-axis.
The ends of the minor axis (often called co-vertices) are at distance
sqrt%28284%2F18%29=sqrt%28142%2F9%29=sqrt%28142%29%2Fsqrt%289%29=sqrt%28142%29%2F3
That distance is called the semi-minor axis.

An ellipse has two highlight%28foci%29 located on the major axis, between the center and the vertices, at a distance from the center c called the focal distance. That distance c, and the semi-minor axis are the legs of a right triangle with the semi-major axis for a hypotenuse.
Applying the Pythagorean theorem, we find that
c%5E2%2B142%2F9=71 --> c%5E2=71-142%2F9 --> c%5E2=71-142%2F9 --> c%5E2=71-142%2F9 --> c%5E2=497%2F9 --> c=sqrt%28497%2F9%29 --> highlight%28c=sqrt%28497%29%2F3%29
So the highlight%28foci%29 are at
(3,-2-sqrt%28497%29%2F3) and (3,-2%2Bsqrt%28497%29%2F3)