Question 830830
<pre>
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 

{{{Ax^2+Bxy+Cy^2+Dx+Ey+F}}}{{{""=""}}}{{{"0"}}}

{{{10x^2 +24xy + 17 y^2 -9}}}{{{""=""}}}{{{"0"}}}

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 {{{theta}}} where

{{{tan(2theta)}}}{{{""=""}}}{{{B/(A-C)}}}{{{""=""}}}{{{24/(10-17)}}}{{{""=""}}}{{{-24/7}}}

{{{tan(2theta)}}}{{{""=""}}}{{{2tan(theta)/(1-tan^2(theta))}}}

{{{-24/7}}}{{{""=""}}}{{{2tan(theta)/(1-tan^2(theta))}}}

Let {{{t=tan(theta)}}}, then

{{{-24/7}}}{{{""=""}}}{{{2t/(1-t^2)}}}

Divide both sides by 2

{{{-12/7}}}{{{""=""}}}{{{t/(1-t^2)}}}


{{{-12(1-t^2)}}}{{{""=""}}}{{{7t}}}

{{{-12+12t^2}}}{{{""=""}}}{{{7t}}}

{{{12t^2-7t-12}}}{{{""=""}}}{{{"0"}}}

{{{(3t-4)(4t+3)}}}{{{""=""}}}{{{"0"}}}

{{{t=4/3}}};  {{{t=-3/4}}}

We can choose either angle.  I'll choose the positive one.

{{{tan(theta)=4/3}}} 
                               
That's a 3-4-5 right triangle {{{drawing(80,100,-.5,3.5,-.5,4.5,triangle(0,0,3,0,3,4),locate(.6,.75,theta),locate(1.5,0.04,3),locate(3.1,2.1,4),
locate(.8,2.4,5)
)}}}  
so {{{sin(theta)=4/5}}},{{{cos(theta)=3/5}}}

We make the substitution 
{{{x=X*cos(theta)-Y*sin(theta)}}}, {{{y=Y*cos(theta)+X*sin(theta)}}}

{{{x=expr(3/5)X-expr(4/5)Y=(3X-4Y)/5}}}, {{{y=expr(3/5)Y+expr(4/5)X=(3Y+4X)/5}}}

or

{{{x=(3X-4Y)/5}}}, {{{y=(3Y+4X)/5}}}

[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

{{{10x^2 +24xy + 17 y^2 -9}}}{{{""=""}}}{{{"0"}}}

{{{10((3X-4Y)/5)^2 +24((3X-4Y)/5)((3Y+4X)/5) + 17((3Y+4X)/5)^2 -9}}}{{{""=""}}}{{{"0"}}}

{{{10(3X-4Y)^2/25 +24(3X-4Y)(3Y+4X)/25 + 17(3Y+4X)^2/25 -9}}}{{{""=""}}}{{{"0"}}}

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

{{{26X^2/9+Y^2/9=1}}}

Divide numerator and denominator of first term by 26

{{{X^2/(9/26)+Y^2/9=1}}}

{{{drawing(400,400,-4,4,-4,4,graph(400,400,-4,4,-4,4,(-sqrt(153-25x^2)-12x)/17),
graph(400,400,-4,4,-4,4,(sqrt(153-25x^2)-12x)/17),line(20,-15,-12,9),
line(-12,-16,15,20),locate(3,4,X),locate(-4,3.3,Y),
line(.74,-.68,.86,-.52),line(1.54,-1.28,1.66,-1.12),
line(2.34,-1.88,2.46,-1.72),line(3.14,-2.48,3.26,-2.32),
line(3.94,-3.08,4.06,-2.92), line(-.86,.52,-.74,.68),
line(-2.46,1.72,-2.34,1.88),
line(-1.66,1.12,-1.54,1.28), line(-3.26,2.32,-3.14,2.48),
line(-4.06,2.92,-3.94,3.08),


line(.68,.74,.52,.86),line(1.28,1.54,1.12,1.66),
line(1.88,2.34,1.72,2.46),line(2.48,3.14,2.32,3.26),
line(3.08,3.94,2.92,4.06),

line(-.68,-.74,-.52,-.86),line(-1.28,-1.54,-1.12,-1.66),
line(-1.88,-2.34,-1.72,-2.46),line(-2.48,-3.14,-2.32,-3.26),
line(-3.08,-3.94,-2.92,-4.06)



)}}}

The center is the origin, because

{{{X^2/(9/26)+Y^2/9=1}}}

can be written:

{{{(X-0)^2/(9/26)+(Y-0)^2/9=1}}}, and is of the form

{{{(X-h)^2/b^2+(Y-k)^2/a^2=1}}} since 9=a²>b²={{{9/26}}}, 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

{{{x=(3X-4Y)/5}}}, {{{y=(3Y+4X)/5}}}
{{{x=(3(0)-4(3))/5}}}, {{{y=(3(3)+4(0))/5}}}
{{{x=-12/5}}}, {{{y=9/5}}}

The upper left vertex is (x,y) = ({{{-12/5}}},{{{9/5}}})

By symmetry, the lower right vertex is ({{{12/5}}},{{{-9/5}}})

To find the foci, we need the value c,

c² = a²-b²
c² = 9-{{{9/26}}}
c² = {{{234/26-9/26}}}
c² = {{{225/26}}}
 c = {{{sqrt(225/26)}}}
 c = {{{sqrt(225)/sqrt(26)}}}
 c = {{{15/sqrt(26)}}}

So the foci on the Y-axis are (X,Y) = (0,{{{"" +- 15/sqrt(26)}}})

We translate the one with the positive Y to its (x,y) coordinates, by using

{{{x=(3X-4Y)/5}}},
{{{x=(3(0)-4(15/sqrt(26)))/5=(-60/sqrt(26))/5=(-60/sqrt(26))*(1/5)=-12/sqrt(26)}}}

{{{y=(3Y+4X)/5}}}
{{{y=(3(15/sqrt(26))+4(0))/5=(45/sqrt(26))/5=(45/sqrt(26))*(1/5)=9/sqrt(26)}}}

The upper left focus is (x,y) = ({{{-12/sqrt(26)}}},{{{9/sqrt(26)}}})

By symmetry, the lower right focus is ({{{12/sqrt(26)}}},{{{-9/sqrt(26)}}})

------------------

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</pre>