For hyperbola `(x^(2))/(cos^(2)alpha)-(y^(2))/(sin^(2)beta)=1` which of the following remains constant with change in `alpha`

Options

A. abscissae of vertex
B. abscissae of foci
C. eccentricity
D. directrix

Explanation

Let `e` be the eccentricity of the hyperbola. Then,
`e^(2)=1+(sin^(2)alpha)/(cos^(2)alpha)=(1)/(cos^(2)alpha)`
Clearly, `e` varies with `alpha`.
Now, `e^(2)=(1)/(cos^(2)alpha)impliese cosalpha=+-1`
Thus, the coordinates of the foci are `(+-1,0)`.
Clearly, abscissae of foci remain constant.
If the centre of the hyperbola is at the point `(h,k)` and the directions of the axes are parallel to the coordinate axes, then its equation is
`((x-h)^(2))/(a^(2))-((y-k)^(2))/(b^(2))=1`
By shifting the origin at `(h,k)` without rotating the coordinate axes, the above equation reduces to
`(X^(2))/(a^(2))-(Y^(2))/(b^(2))=1`, where `x=X+h` and `y=Y+k`.
REMARK As for the ellipse, if a point `P(x,y)` moves in the plane of two perpendicular straight lines `a_(1)x+b_(1)y+c_(1)=0` and `b_(1)x-a_(1)y+c_(2)=0` in such a way that
`(((a_(1)x+b_(1)y+c_(1))/(sqrt(a_(1)^(2)+b_(1)^(2)))))/(a^(2))-(((b_(1)x-a_(1)y+c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)))))/(b^(2))=1`,
then the locus of `P` is a hyperbola whose transverse axis lies along `b_(1)x-a_(1)y+c_(2)=0` and conjugate axis along the line `a_(1)x+b_(1)y+c_(1)=0`. The lengths of transverse and conjugate axes are `2a` and `2b` respectively.