Question
For the Hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1,$$ which of the following remains constant when $$\alpha $$ varies $$=?$$
A.
abscissae of vertices
B.
abscissae of foci
C.
eccentricity
D.
directrix
Answer :
abscissae of foci
Solution :
Given, equation of hyperbola is $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1$$
We know that the equation of hyperbola is $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$
We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$
$$\eqalign{
& \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right) \cr
& \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2} \cr
& \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \cr
& \Rightarrow e = \sec \,\alpha \cr
& \therefore ae = \cos \,\alpha .\frac{1}{{\cos \,\alpha }} = 1 \cr} $$
Co-ordinates of foci are ($$\left( { \pm ae,\,0} \right)$$ i.e., $$\left( { \pm 1,\,0} \right)$$
Hence, abscissae of foci remain constant when $$\alpha $$ varies.