A hyperbola has the points 112 43 and 62 What is its equatio

A hyperbola has the points (1,12) (4,3) and (6,2). What is its equation?

Solution

We\'ll start with the following equation for hyperbola:

y= (a+ bx)/(c+dx)

Dividing numerator and denominator of RHS by \'d\'. We get:

y= (a/d + bx/d)/ (c/d +x)

Let a/d= A, b/d = B and c/d= C

So the new equation is:

y= (A + Bx)/ (C + x)

Putting the values of each point in the above equation, one by one, we get three equations:

A+ B-12C= 12...(i)

A+ 6B- 2C= 12 ...(ii)

A+4B-3C = 12...(iii)

Solving (i) and (ii), we get:

B + 2C= 0...(iv)

Solving (ii) and (iii), we get:

2B+ C= 0...(v)

Solving (iv) and (v), we get:

B=0 and C=0...(vi)

From (i) and (vi), we get:

A=12

Putting the values of A, B, C in y= (A+Bx)/(C+x), we get:

y= 12/x or xy=12, which is the required equation.

Lets take d=1

This reduces the equation to:

y= (a+bx)/ (c+x)