The tangents drawn from a point P to ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` make angles `alpha and beta` with the transverse axis of the hyperbola, then

Explanation

The equation of any tangent to
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1" "(1)`
is given by
`y=mx+sqrt(a^(2)m^(2)+b^(2))`

Sinc the passes through P(h,k) we have
`k=mh+sqrt(a^(2)m^(2)+b^(2))`
or `m^(2)(h^()-a^(12))-2kmnh++(k^(2)+b^(2))=0" "(2)`
As (2) is quadratic in m, having two roots `m_(1) andm_(2)` (ay), we age
`m_(1)+m_(2)=(2hk)/(h^(2)-a^(2)),m_(1)m_(2)=(k^(2)-b^(2))/(h^(2)-a^(2)) " "(3)`
or `tan(alpha+beta)=(tan alpha+tanbeta)/(1-tan alpha tan beta)`
`=(m_(1)+m_(2))/(1-m_(1)m_(2))=(2hk//(h^(2)-a^(2)))/(1-{(kl^(2)-b^(2))//(h^(2)-a^(2))})=(2hk)/(h^(2)-k^(2)-a^(2)+b^(2))`
(a) `alpha+beta=(cpi)/(2)` ltbrtgt When c even,
`m_(1)+m_(2)=0`
`(2kh)/(h^(2)-a^(2))=0`
or `2kh=0`
`xy=0`
Which is the equation of a pair of straight lines.
Whnen c is odd,
`1-m_(1)m_(2)=0`
`or (k^(2)-b^(2))/(h^(2)-a^(2))=1`
Therefore, the locus of (h,k) is `y^(2)-b^(2)=x^(2)-a^(2)`
which is a hyperbola.
(b) `m_(1)m_(2)=c`
or `(k^(2)-b^(2))/(h^(2)-a^(2))=c`
when `c=0,k=+-b`, the locus is a pair of straigth lines.
When `c=1,h^(2)-k^(2)=a^(2)-b^(2)`, the locusb is a hyperbola.
When `c-1,h^(2)+k^(2)=a^(2)+b^(2),` the locus is a circle
When `c=-2,2h^(2)+k^(2)=2a^(2)+b^(2)`, the locus is an ellipse
(c) `tan alpha+beta=c`
or `m_(1)+m_(2)=c`
When `c=0,kh,2-kh=0`, the locus ia pair of straight lines,
When `c ne 0`
`c(h^(2)-a^(2))-2kh=0`
the locus of (h,k) is `cx^(2)-2kh=0`
`Deta=-ca^(2)ne0`
Also,` h^(2)-ab=gt0`
Therefore, the locus is a hyperboal or `cne=0`
(d) `cot alpha+cos beta=c`
or `(1)/(m_(1))+(1)/(m_(2))=c`
or `(m_(1)+m_(2))/(m_(1)m_(2))=c`
or `(2kh)/(k^(2)-b^(2))=c`
When c=0, the locus is pair of straigth lines
Wehn `cne0`, the locus is hyperbola (as in previous case c)