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  •  Given <img alt="\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B%5Cmathrm%7Bx%7D%7D%7B%5Cmathrm%7Ba%7D%7D&plu

 Given \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1 and \mathrm{ax}+\mathrm{by}=1 are two variable lines,  \mathrm{'a'} and \mathrm{'b'} being the parameters connected by the relation \mathrm{a^{2}+b^{2}=a b}.The locus of the point of intersection has the equation

Option: 1

\mathrm{x^{2}+y^{2}+x y-1=0}


Option: 2

\mathrm{x^{2}+y^{2}-x y+1=0}


Option: 3

\mathrm{x^{2}+y^{2}+x y+1=0}


Option: 4

\mathrm{x^{2}+y^{2}-x y-1=0}


Answers (1)

best_answer

Let (h, k) be point of intersection then

\begin{array}{lll} \frac{\mathrm{h}}{\mathrm{a}}+\frac{\mathrm{k}}{\mathrm{b}}=1 & \text { given } & \mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{ab} \\ \mathrm{ah}+\mathrm{kb}=1 & \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=1 \end{array}
---------------------
\mathrm{multiply \: \mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{hk}\left(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}\right)=1}
                     \mathrm{\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{hk}=1}
                     \mathrm{\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{xy}-1=0}

Note that the locus is not physically viable

Posted by

jitender.kumar

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