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 \[SP=a(1-ecos\theta )\] |
| \[\therefore \frac{2a}{SP}-1=\frac{2a}{a(1-\,\,e\,\,cos\theta )}-1=\frac{1+e\cos \theta }{1-e\cos \theta }=\frac{{{b}^{2}}}{{{p}^{2}}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
