Question
If $$p$$ is the length of the perpendicular from 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 $$\frac{{2a}}{{SP}} - 1 = ?$$
A.
$$\frac{{{a^2}}}{{{p^2}}}$$
B.
$$\frac{{{b^2}}}{{{p^2}}}$$
C.
$${p^2}$$
D.
$$\frac{{{a^2} + {b^2}}}{{{p^2}}}$$
Answer :
$$\frac{{{b^2}}}{{{p^2}}}$$
Solution :
Let the point $$P$$ be $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$
The tangent at $$P$$ is $$\frac{x}{a}\cos \,\theta + \frac{y}{b}\sin \,\theta = 1......\left( {\text{i}} \right)$$
The perpendicular distance $$p$$ of $$S\left( {ae,\,0} \right)$$ form $$\left( {\text{i}} \right)$$ is given by
$$\eqalign{
& {p^2} = \frac{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}}{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}} \cr
& \Rightarrow \frac{1}{{{p^2}}} = \frac{{\frac{{{{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\theta }}{{{b^2}}}}}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr
& \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{\frac{{{b^2}}}{{{a^2}}}{{\cos }^2}\theta + 1 - {{\cos }^2}\theta }}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr
& \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{\left( {\frac{{{b^2}}}{{{a^2}}} - 1} \right){{\cos }^2}\theta + 1}}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr
& \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{1 - {e^2}{{\cos }^2}\theta }}{{{{\left( {e\,\cos \,\theta - 1} \right)}^2}}} \cr
& \Rightarrow \frac{{{b^2}}}{{{p^2}}} = \frac{{1 + e\,\cos \,\theta }}{{1 - e\,\cos \,\theta }} \cr
& {\text{Now, }}SP = a\left( {1 - e\,\cos \,\theta } \right) \cr
& \therefore \,\frac{{2a}}{{SP}} - 1 = \frac{{2a}}{{a\left( {1 - e\,\cos \,\theta } \right)}} - 1 = \frac{{1 + e\,\cos \,\theta }}{{1 - e\,\cos \,\theta }} = \frac{{{b^2}}}{{{p^2}}} \cr} $$