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Question
A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA − PB = k (k ≠ 0), then the locus of P is
Options
a hyperbola
a branch of the hyperbola
a parabola
an ellipse
MCQ
Solution
a hyperbola
\[\text { Let } P(x, y) \text { be any point on the hyperbola } \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 . \]
By definition, we have:
\[PA = e\left( x - \frac{a}{e} \right) = ex - a\]
\[\text { and } PB = e\left( x + \frac{a}{e} \right) = ex + a\]
\[ \therefore PB - PA = \left( ex + a \right) - \left( ex - a \right) = 2a = k\]
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Introduction of Hyperbola
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