# Normal to the hyperbola Question

1. Jul 24, 2007

1. The problem statement, all variables and given/known data
The normal to the hyperbola (x^2)/2 - y^2 = 1 at P (sqrt 3, sqrt 0.5) cuts the y-axis at A and the x-axis at B. Show that PA:PB = 2:1

2. Relevant equations

Equation of normal to general hyperbola at (x1,y1) is x(a^2)/x1 + y(b^2)/y1 = a^2 + b^2

3. The attempt at a solution

Okay so that makes the eqn of this particular normal: 2x/sqrt 3 + (sqrt 2)y = 3
So A [0, (sqrt 3)/2] and B [3(sqrt 3)/2, 0]. So using the internal division of lines formula given the ratio 2:1 to prove that the point which cuts the line in that particular ratio is P, I end up with:

x = [1*0 + 2*3(sqrt3)/2] / (1+2)
= sqrt 3
y = [1*(sqrt 3)/2 + 2*0] / (1+2)
= 1/(2sqrt 3)

Which is not P... any insights into where I went wrong?

2. Jul 24, 2007

### chaoseverlasting

Are you sure your equations are right? Maybe Im wrong, but shouldnt it be a^2-b^2 on the RHS?

3. Jul 24, 2007

Thanks for replying! I'm pretty sure it is a^2 + b^2 though, the equation was given in the question before and I quickly derived it to double check.

4. Jul 25, 2007

### chanvincent

You have made a major algebraic error in finding to coordination of A and B. I notice A, B and P is not colinear

5. Jul 25, 2007

Sorry, I'm a bit confused. How come A,B and P aren't collinear? Aren't they on the same line (the normal to the hyperbola?)

6. Jul 25, 2007

### chanvincent

Yes... I meant in your solution, A $$(0,\sqrt{3}/2)$$, B $$(3\sqrt{3}/2, 0)$$, and P$$(\sqrt{3}, \sqrt{0.5})$$ is not collinear..

7. Jul 25, 2007