Is the curve $(x^{2} - a^{2}) (y^{2} - b^{2}) + 2 x y + 3 x + 4 y = 7$ $(x^2-a^2)(y^2-b^2)+2xy+3x+4y=7$ has only 2 asymptotes parallel to coordinate axes?

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Is the curve $(x^{2} - a^{2}) (y^{2} - b^{2}) + 2 x y + 3 x + 4 y = 7$ $(x^2-a^2)(y^2-b^2)+2xy+3x+4y=7$ has only 2 asymptotes parallel to coordinate axes?

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Answer 1 · 2017-05-25T14:27:53

See below.

Explanation:

Solving for $x$ $x$ (using the Bashkara formula) we have

$x = \frac{\pm (3 + 2 y) + \sqrt{{(3 + 2 y)}^{2} - 4 (y^{2} - b^{2}) (a^{2} b^{2} + 4 y - a^{2} y^{2} - 7)}}{2 (y^{2} - b^{2})}$ $x = (pm(3 +2 y) + sqrt[(3 + 2 y)^2 - 4 (y^2-b^2) (a^2 b^2 + 4 y - a^2 y^2-7)])/( 2 (y^2-b^2))$

and solving for $y$ $y$ we have

$y = \frac{\pm (4 + 2 x) + \sqrt{{(4 + 2 x)}^{2} - 4 (x^{2} - a^{2}) (a^{2} b^{2} + 3 x - b^{2} x^{2} - 7)}}{2 (x^{2} - a^{2})}$ $y=(pm(4 + 2 x) + sqrt[(4 + 2 x)^2 - 4 (x^2-a^2) (a^2 b^2 + 3 x - b^2 x^2-7)])/(2 (x^2-a^2))$

so we have four asymptotes:

$y = \pm b$ $y = pm b$ and

$x = \pm a$ $x = pm a$

Attached a plot for $a = 1$ $a=1$ , $b = 2$ $b=2$

enter image source here

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Is the curve $(x^{2} - a^{2}) (y^{2} - b^{2}) + 2 x y + 3 x + 4 y = 7$ $(x^2-a^2)(y^2-b^2)+2xy+3x+4y=7$ has only 2 asymptotes parallel to coordinate axes?

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Explanation:

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Is the curve (x^2-a^2)(y^2-b^2)+2xy+3x+4y=7(x2−a2)(y2−b2)+2xy+3x+4y=7(x^2-a^2)(y^2-b^2)+2xy+3x+4y=7 has only 2 asymptotes parallel to coordinate axes?

1 Answer

Explanation:

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Impact of this question

Is the curve $(x^{2} - a^{2}) (y^{2} - b^{2}) + 2 x y + 3 x + 4 y = 7$ $(x^2-a^2)(y^2-b^2)+2xy+3x+4y=7$ has only 2 asymptotes parallel to coordinate axes?