A) \[ap=rp+1\]
B) \[rp=ap\]
C) \[ap=rp+1\]
D) \[ap=rp\]
Correct Answer: D
Solution :
Tangent to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]at\[(a\cos \theta ,\,\,b\sin \theta )\]is \[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] ... (i) \[\therefore \]\[p=\]perpendicular distance from focus\[(ae,\,\,0)\]to the line (i) \[=\frac{\left| \frac{ae}{a}\cos \theta +0-1 \right|}{\sqrt{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}}\] \[=\frac{1-e\cos \theta }{\sqrt{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}}\] ? (i) Also, \[p=\]perpendicular distance from centre (0, 0) to the line (i) \[=\frac{1}{\sqrt{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}}\] ? (ii) Again, \[r=SP=a(1-e\cos \theta )\] \[\therefore \] \[ap=\frac{a-ae\cos \theta }{\sqrt{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}}=rp\]
You need to login to perform this action.
You will be redirected in
3 sec
