Math, asked by Missx20, 4 months ago

Eliminate theta from x = a + b sec theta , y = c + d tan theta

Answers

Answered by MaheswariS
10

\textbf{Given:}

\mathsf{x=a+b\;sec\theta}

\mathsf{y=c+d\;tan\theta}

\textbf{To find:}

\textsf{An equation obtained by eliminating}\;\theta

\textbf{Solution:}

\textsf{Consider,}

\mathsf{x=a+b\;sec\theta}

\mathsf{x-a=b\;sec\theta}

\implies\mathsf{\dfrac{x-a}{b}=sec\theta}

\mathsf{and}

\mathsf{y=c+d\;sec\theta}

\mathsf{y-c=d\;sec\theta}

\implies\mathsf{\dfrac{y-c}{d}=sec\theta}

\mathsf{We know that,}

\mathsf{sec^2\theta-tan^2\theta=1}

\mathsf{\left(\dfrac{x-a}{b}\right)^2-\left(\dfrac{y-c}{d}\right)^2=1}

\implies\boxed{\mathsf{\dfrac{(x-a)^2}{b^2}-\dfrac{(y-c)^2}{d^2}=1}}

\textbf{Find more:}

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Answered by mahek77777
11

\textbf{Given:}

\mathsf{x=a+b\;sec\theta}

\mathsf{y=c+d\;tan\theta}

\textbf{To find:}

\textsf{An equation obtained by eliminating}\;\theta

\textbf{Solution:}

\textsf{Consider,}

\mathsf{x=a+b\;sec\theta}

\mathsf{x-a=b\;sec\theta}

\implies\mathsf{\dfrac{x-a}{b}=sec\theta}

\mathsf{and}

\mathsf{y=c+d\;sec\theta}

\mathsf{y-c=d\;sec\theta}

\implies\mathsf{\dfrac{y-c}{d}=sec\theta}

\mathsf{We know that,}

\mathsf{sec^2\theta-tan^2\theta=1}

\mathsf\red{\left(\dfrac{x-a}{b}\right)^2-\left(\dfrac{y-c}{d}\right)^2=1}

\implies\boxed{\mathsf\red{\dfrac{(x-a)^2}{b^2}-\dfrac{(y-c)^2}{d^2}=1}}

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