Some problems with conic sections

[Mspike6]

Here are some conic questions that am having problems with :

1) The general form of a particular ellipse is show Below. If the conic is translated 2 units left and 1 unit down, determine the new general term.

2x²+y²-2x+3y-9=0

Solution,
i figureed that we will have to convert this to Sandard form so we can apply the Trans;ations.

2x²-2x y²-3y=0

Then we need to complete the squares

2(x²-x+[tex]\frac{1}{4}[/tex]) + (y²-3y+[tex]\frac{9}{4}[/tex])=9+1/2+9/4

When i divided the whole thing by 11.75 so i can get 1 on the Right hand side, a fractions come up on the right hand side, and i just can't get it to the Standard form of the elipse

2) Find the equation of a hyperbola in standard form or general form, that has its centre at (-2, 5), one vertex at (-2, 10), and the slope of one of its asymptotes is 5/4 .

Solution:
I drew the Hyperbola so i can imagine it, and i guessed that a = 5 since the center point is at (-2,5) and the Vertex is at (-2,10)

The slops are b/a and -b/a
So it should be b/5 not b/4

aint what am saying right ?

 

[rl.bhat]

The standard equation of ellipse becomes
(x - 1/2)^2/(47/8) + (y-3/2)^2/(47/4) = 1

 

[chaoseverlasting]

Whats the question?

 

[Mspike6]

Sorry, i meant to ask if what i did is correct , or is there something am missing ?

 

[HallsofIvy]

Here are some conic questions that am having problems with :

1) The general form of a particular ellipse is show Below. If the conic is translated 2 units left and 1 unit down, determine the new general term.

2x²+y²-2x+3y-9=0

There is no need to convert to standard form. "Translating 2 units left" is exactly the same as adding 2 to x. "Translating one unit down" is exactly the same as subtracting 1 from y. Replace x by x+2 and y by y- 1.

2) Find the equation of a hyperbola in standard form or general form, that has its centre at (-2, 5), one vertex at (-2, 10), and the slope of one of its asymptotes is 5/4 .

Rotating through an angle [itex]\theta[/itex] gives [itex]x'= x cos(\theta)+ y sin(\theta)[/itex], [itex]y'= x sin(\theta)- y cos(\theta)[/itex]. Replace x by [itex]x cos(\theta)+ y sin(\theta)[/itex], y by [itex]x sin(\theta)- y cos(\theta)[/itex]. [itex]\theta= arctan(5/4)[/itex].

 

[Mspike6]

There is no need to convert to standard form. "Translating 2 units left" is exactly the same as adding 2 to x. "Translating one unit down" is exactly the same as subtracting 1 from y. Replace x by x+2 and y by y- 1.



Rotating through an angle [itex]\theta[/itex] gives [itex]x'= x cos(\theta)+ y sin(\theta)[/itex], [itex]y'= x sin(\theta)- y cos(\theta)[/itex]. Replace x by [itex]x cos(\theta)+ y sin(\theta)[/itex], y by [itex]x sin(\theta)- y cos(\theta)[/itex]. [itex]\theta= arctan(5/4)[/itex].

Thanks a lot ! :D
really appreciate it