How do you write $x^2 - 8y^2 - 14x -15 = 0$ in standard form?

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Answer 1 · 2015-05-21T14:57:29

This is a hyperbola with a horizontal transverse axis so its standard form is

$(x-h)^2/a^2 - (y-k)^2/b^2 = 1$

Given $x^2-8y^2-14x-15 = 0$

$x^2-14x+49 - 8y^2 = 15 +49$

$(x-7)^2-8y^2 = 64$

$(x-7)^2/8^2 - y^2/(2sqrt(2))^2 = 1$

How do you write x^2 - 8y^2 - 14x -15 = 0x^2 - 8y^2 - 14x -15 = 0 in standard form?