A) \[\frac{{{a}^{2}}}{{{x}^{2}}}-\frac{{{b}^{2}}}{{{y}^{2}}}=1\]
B) \[\frac{{{a}^{2}}}{{{x}^{2}}}+\frac{{{b}^{2}}}{{{y}^{2}}}=1\]
C) \[\frac{{{a}^{2}}}{{{y}^{2}}}-\frac{{{b}^{2}}}{{{x}^{2}}}=1\]
D) None of these
Correct Answer: A
Solution :
[a] Eq. of the tangent at the point \['\theta '\] is |
\[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\] |
\[\Rightarrow \] A is \[(a\,cos\theta ,\,\,0)\] and B is \[(0,\,\,-b\,cot\,\,\theta )\] |
Let P be \[(h,k)\Rightarrow h=acos\theta ,k=-bcot\theta \] |
\[\Rightarrow \frac{k}{h}=-\frac{b}{a\sin \theta }\Rightarrow \sin \theta =\frac{bh}{ak}\] and \[\cos \theta =\frac{h}{a}.\] |
Square and add, |
\[\Rightarrow \frac{{{b}^{2}}{{h}^{2}}}{{{a}^{2}}{{k}^{2}}}+\frac{{{h}^{2}}}{{{a}^{2}}}=1\Rightarrow \frac{{{b}^{2}}}{{{k}^{2}}}+1=\frac{{{a}^{2}}}{{{h}^{2}}}\] |
\[\Rightarrow \frac{{{a}^{2}}}{{{h}^{2}}}-\frac{{{b}^{2}}}{{{k}^{2}}}=1\] |
Hence, locus of P is \[\frac{{{a}^{2}}}{{{x}^{2}}}-\frac{{{b}^{2}}}{{{y}^{2}}}=1\] |
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