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# (((x^2)/2)-(x/sqrt 2)-(y^2/2)-(3y/sqrt 2) +3=0 in standard form of hyperbola

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(((x^2)/2)-(x/sqrt 2)-(y^2/2)-(3y/sqrt 2) +3=0 in standard form of hyperbola

Guest May 29, 2015

#2
+89953
+5

This is a hyperbola....the "minus" sign between the first two terms tells us this.. it also opens "upward and downward"........this is easy to spot because the variable that comes first - in this case, "y' - will tell us which axis the hyperbola intersects

As in all conic forms, the center is found at (1/√2, -3/√2)

The transverse axis is given by the equation ....x = 1/√2.......the focal points lie on this line and are located at (1/√2, -3/√2 ± √[a2 + b2]) =  (1/√2, -3/√2 ± 10√2)

And the asymptotes are given by ....  y = ±(a/b)(x - h) + k  =

±(10/10)(x - 1/√2) - 3/√2  = ± (x - 1/√2) - 3/√2

Here's a graph that puts all this together.......https://www.desmos.com/calculator/pskcr0hrmy

Here's a good website covering the basics of the hyperbola.....http://www.purplemath.com/modules/hyperbola.htm

CPhill  May 29, 2015
#1
+93656
+5

$$\\\frac{x^2}{2}-\frac{x}{\sqrt2}-\frac{y^2}{2}-\frac{3y}{\sqrt 2} +3=0\\\\ x^2-\sqrt2\;x-y^2-3\sqrt2\;y +6=0\\\\ \left[x^2-\sqrt2\;x+(\frac{\sqrt2}{2})^2\right]-\left[y^2+3\sqrt2\;y+(\frac{3\sqrt2}{2})^2\right]=-6+(\frac{\sqrt2}{2})^2-(\frac{3\sqrt2}{2})^2\right\\\\ \left[x^2-\sqrt2\;x+\frac{1}{2}\right]-\left[y^2+3\sqrt2\;y+\frac{9}{2}\right]=-6+\frac{1}{2}-\frac{9}{2}\\\\ \left[x^2-\sqrt2\;x+\frac{1}{2}\right]-\left[y^2+3\sqrt2\;y+\frac{9}{2}\right]=-10\\\\ \left[(x-\frac{1}{\sqrt2})^2\right]-\left[(y+\frac{3}{\sqrt2})^2\right]=-10\\\\ \left[\frac{(y+\frac{3}{\sqrt2})^2}{10}\right]-\left[\frac{(x-\frac{1}{\sqrt2})^2}{10}\right]=1\\\\$$

Here is the graph

https://www.desmos.com/calculator/r03pjfgkxs

Melody  May 29, 2015
#2
+89953
+5

This is a hyperbola....the "minus" sign between the first two terms tells us this.. it also opens "upward and downward"........this is easy to spot because the variable that comes first - in this case, "y' - will tell us which axis the hyperbola intersects

As in all conic forms, the center is found at (1/√2, -3/√2)

The transverse axis is given by the equation ....x = 1/√2.......the focal points lie on this line and are located at (1/√2, -3/√2 ± √[a2 + b2]) =  (1/√2, -3/√2 ± 10√2)

And the asymptotes are given by ....  y = ±(a/b)(x - h) + k  =

±(10/10)(x - 1/√2) - 3/√2  = ± (x - 1/√2) - 3/√2

Here's a graph that puts all this together.......https://www.desmos.com/calculator/pskcr0hrmy

Here's a good website covering the basics of the hyperbola.....http://www.purplemath.com/modules/hyperbola.htm

CPhill  May 29, 2015
#3
+93656
0

Thanks Chris, that is a really great addition!

Melody  May 29, 2015