What is 1/(a+b)+1/(a-b)= 2/(a-1) ?

The solution to 1/(a+b)+1/(a-b)= 2/(a-1) is Parabola with (h,k)=(0,0),p= 1/4

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1/(a+b)+1/(a-b)= 2/(a-1)

\frac{1}{a + b} + \frac{1}{a - b} = \frac{2}{a - 1}

(h, k) = (0, 0), p = \frac{1}{4}

Solution steps

\frac{1}{a + b} + \frac{1}{a - b} = \frac{2}{a - 1}

Rewrite

\frac{1}{a + b} + \frac{1}{a - b} - \frac{2}{a - 1} = 0

in the standard form

4 \cdot \frac{1}{4} (a - 0) = (b - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{4}

x = \frac{1}{4} y^{2}

9 y^{2} + 12 x = 0

y^{2} + 6 x + 10 y + 19 = 0

y^{2} + 12 x = 0

- y^{2} = 6 y + x

What is 1/(a+b)+1/(a-b)= 2/(a-1) ?
The solution to 1/(a+b)+1/(a-b)= 2/(a-1) is Parabola with (h,k)=(0,0),p= 1/4