Question

MEDAL AND FAN!!
A hyperbola with a horizontal transverse axis contains the point at (4, 3). The equations of the asymptotes are y - x = 1 and y + x = 1.
Write an equation for the hyperbola.

Answer 1

@ganeshie8 please help

Answer 2

@math-geek @hartnn help

Answer 3

@IMStuck can u help me

Answer 4

@AravindG

Answer 5

Give me a sec to work on this one...hyperbolas are stinkers! ; )

Answer 6

sure thanks in advance

Answer 7

umm...u there???

Answer 8

I am and I am really struggling to find the a and the b here.

Answer 9

The best I can do is this:\[\frac{ x ^{2} }{ 1 }-\frac{ (y-1)^{2} }{ 1 }=1\]

Answer 10

okay thnx

Answer 11

ACtually I think I DID get it...\[\frac{ x ^{2} }{ 4 }-\frac{ (y-1)^{2} }{ 3 }=1\]

Answer 12

Can you check it?

Answer 13

i checked it
it is incorrect
but i got it now so nvr mind

Answer 14

what is it!? Please tell me so I can find out how to find a and b. Please!

Answer 15

http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections.faq.question.881602.html

Answer 16

its very similar to my ques

Answer 17

OMG thank you for that website! I love these informational sites; they are such a great help!

Answer 18

ur welcum

Answer 19

We're given y-x=1 and y+x=1, which are lines that intersect at (0,1) at 90 degrees.
This means that the "rectangle" is a square, or a=b.
c (distance of the centre to the foci) is therefore \(\sqrt2 a\).
The general equation of a hyperbola with centre (h,k) is given by:
\(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\) since the hyperbola has a horizontal axis.
The conic has a centre of (0,1), so the equation reduces to:
\(\frac{x^2}{a^2}+\frac{(y-1)^2}{b^2}=1\)
In addition, from the angles of the two asymptotes, we further deduced that a=b.
The equation is therefore:
\(\frac{x^2}{a^2}+\frac{(y-1)^2}{a^2}=1\)
Now we'll substitute (4,3) into the equation to solve for a.
\(\frac{4^2}{a^2}+\frac{(3-1)^2}{a^2}=1\)
=>
\(a^2=16-4=12\)
or \(a^2=12\)
The final equation is therefore:
\(\frac{x^2}{12}+\frac{(y-1)^2}{12}=1\)