Since the other tutor gave you such a pitiful answer, I thought
I'd take the trouble to give you what you wanted. It's quite a
tough problem.
The general form of a conic is
A=10, B=24, C=17
Discriminant B²-4AC = 24²-4(10)(17) = -104
This being negative tell us that the equation represents an
ellipse.
We will rotate the axes through an angle of where
Let , then
Divide both sides by 2
;
We can choose either angle. I'll choose the positive one.
That's a 3-4-5 right triangle
so ,
We make the substitution
,
,
or
,
[Most books use x' and y', but I am using capital X and Y because
primed letters are difficult to work with. So be careful in
the rest of the problem to distinguish between capital X and small x
and capital Y and small y. Remember the capital letters refer to the
rotated axes and small ones refer to the original un-rotated axes.]
Substitute into
Multiply through by 25
10(3X-4Y)²+24(3X-4Y)(3Y+4X)+17(3Y+4X)²-225 = 0
10(9X²-24XY+16Y²)+24(9XY+12X²-12Y²-16XY)+17(9Y²+24XY+16X²)-225 = 0
90X²-240XY+160Y²+24(12X²-7XY-12Y²)+153Y²+408XY+272X²-225 = 0
90X²-240XY+160Y²+288X²-168XY-288Y²+153Y²+408XY+272X²-225 = 0
650X²+25Y²-225=0
Divide through by 25
26X²+Y²-9 = 0
To get that in standard form,
26X²+Y² = 9
Divide through by 9 to get 1 on the right
Divide numerator and denominator of first term by 26
The center is the origin, because
can be written:
, and is of the form
since 9=a²>b²=, and
where (h,k) is the center.
center = (X,Y) = (x,y) = (0,0)
a²=9, so a=3, so the (X,Y) coordinates of the vertex is on the Y-axis
is (X,Y) = (0,3). We translate this to its (x,y) coordinates, by using
,
,
,
The upper left vertex is (x,y) = (,)
By symmetry, the lower right vertex is (,)
To find the foci, we need the value c,
c² = a²-b²
c² = 9-
c² =
c² =
c =
c =
c =
So the foci on the Y-axis are (X,Y) = (0,)
We translate the one with the positive Y to its (x,y) coordinates, by using
,
The upper left focus is (x,y) = (,)
By symmetry, the lower right focus is (,)
------------------
To find the domain and range exactly is really murder.
To find the range exactly
1. Solve the original equation for y using the quadratic formula
2. You will have two functions, one using + and one using -
3. Find their derivatives
4. Set each equal to 0 and solve for x
5. Substitute in the result of 1 to find y in each
6. The range will be [smaller value of y, larger value of y]
To find the domain exactly
1. Interchange x and y in the original equation.
2. Solve the that equation for y using the quadratic formula
3. You will have two functions, one using + and one using -
4. Find their derivatives.
5. Set each equal to 0 and solve for x
6. Substitute in the result of 2 to find y in each
7. The domain will be [smaller value of y, larger value of y]
That'll take many hours. I just used a TI-84 to find the
approximate domain and range,
The approximate domain is [-2.47386337,2.47386337]
The approximate range is [-2.283567,2.283567]
Edwin