Question
For $w=z^{2}, u=x^{2}-y^{2}$ and $v=2 x y .$ If $x y=1, v=2$ and so the hyperbola $x y=1$ is mapped onto the line $v=2.$
Step 1
We are also given that $xy = 1$ and $v = 2$. Show more…
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The equation of the hyperbola is of the form (a) $\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$ (b) $\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$
Analytic Geometry
The Hyperbola
The equation of the hyperbola is of the form $$ \begin{array}{l}{\text { (a) } \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1} \\ {\text { (b) } \frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1} \\ {\text { (c) } \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1} \\ {\text { (d) } \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1}\end{array} $$
Using the figure to the right. The equation of the hyperbola is of the form (a) $\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$ (b) $\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$ (c) $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$ (d) $\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$
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