Question #60158

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Answer 1 · 2017-02-11T18:05:00

${(\frac{x - h}{a})}^{2} + {(\frac{y - k}{b})}^{2} = 1$ $((x-h)/a)^2+((y-k)/b)^2=1$

$\frac{x - h}{a} = cos θ$ $(x-h)/a=costheta$ and $\frac{y - k}{b} = sin θ$ $(y-k)/b=sintheta$

We know that ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$ so

${(\frac{x - h}{a})}^{2} + {(\frac{y - k}{b})}^{2} = 1$ $((x-h)/a)^2+((y-k)/b)^2=1$ which is the equation of an ellipse.