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