The locus of the point of intersection of the straight lines \(\frac{x}{a}+\frac{y}{b}=K\) and \(\frac{x}{a}-\frac{y}{b}=\frac{1}{k}\), where \(k\) is a non-zero real variable, is given by

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The locus of the point of intersection of the straight lines \(\frac{x}{a}+\frac{y}{b}=K\) and \(\frac{x}{a}-\frac{y}{b}=\frac{1}{k}\), where \(k\) is a non-zero real variable, is given by

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kritika · Answer 1 · 2021-12-19T16:58:16+0000

Ans: (D)
Hint: Let the point intersection be \((\alpha, \beta)\).
so, \(\frac{\alpha}{a}+\frac{\beta}{b}=k\) and \(\frac{\alpha}{a}-\frac{\beta}{b}=\frac{1}{k}\)
\(\Rightarrow \frac{\alpha^{2}}{a^{2}}-\frac{\beta^{2}}{b^{2}}=1\)
\(\therefore\) Locus \(: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) which is equation of a hyperbola.

The locus of the point of intersection of the straight lines \(\frac{x}{a}+\frac{y}{b}=K\) and \(\frac{x}{a}-\frac{y}{b}=\frac{1}{k}\), where \(k\) is a non-zero real variable, is given by

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