A) \[\frac{{{a}^{2}}}{{{p}^{2}}}\]
B) \[\frac{{{b}^{2}}}{{{p}^{2}}}\]
C) \[{{p}^{2}}\]
D) \[\frac{{{a}^{2}}+{{b}^{2}}}{{{p}^{2}}}\]
Correct Answer: B
Solution :
[b] Let the point P be \[(a\,cos\,\theta ,\,\,b\,sin\,\theta )\] |
The tangent at P is \[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] ? (i) |
The perpendicular distance P of S (ae, 0) |
Form (i) is given by \[{{p}^{2}}=\frac{{{(ecos\theta -1)}^{2}}}{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}\] |
\[\Rightarrow \frac{1}{{{p}^{2}}}=\frac{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}{{{(ecos\theta -1)}^{2}}}\] |
\[\Rightarrow \frac{{{b}^{2}}}{{{p}^{2}}}=\frac{\frac{{{b}^{2}}}{{{a}^{2}}}{{\cos }^{2}}\theta +1-{{\cos }^{2}}\theta }{{{(ecos\theta -1)}^{2}}}\] |
\[=\frac{\left( \frac{{{b}^{2}}}{{{a}^{2}}}-1 \right){{\cos }^{2}}\theta +1}{{{(ecos\theta -1)}^{2}}}\] |
\[=\frac{1-{{e}^{2}}{{\cos }^{2}}\theta }{{{(ecos\theta -1)}^{2}}}=\frac{1+e\cos \theta }{1-e\cos \theta }\] |
Now \[SP=a(1-ecos\theta )\] |
\[\therefore \frac{2a}{SP}-1=\frac{2a}{a(1-\,\,e\,\,cos\theta )}-1=\frac{1+e\cos \theta }{1-e\cos \theta }=\frac{{{b}^{2}}}{{{p}^{2}}}\] |
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