Conic-Sections Question 542
Question: If p is the length of the perpendicular form the focus S of the ellipse $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ to a tangent at a point P on the ellipse, then $ \frac{2a}{SP}-1= $
Options:
A) $ \frac{a^{2}}{p^{2}} $
B) $ \frac{b^{2}}{p^{2}} $
C) $ p^{2} $
D) $ \frac{a^{2}+b^{2}}{p^{2}} $
Show Answer
Answer:
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{( \frac{b^{2}}{a^{2}}-1 ){{\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-ecos\theta )}-1=\frac{1+e\cos \theta }{1-e\cos \theta }=\frac{b^{2}}{p^{2}} $
BETA
