find the equation of parabola which is symmetric about y-axis,and passes through point (2,-3).

Answer

A, B

Explanation

(1), (2)
Let (h, k) be the excenter. Then,
`h=(ae(ae sec theta+a)-ae(ae sectheta-a)-2ae(a sec theta))/(2ae(sec theta-1))=-a` ltBrgt `"or "x=-a("for "S'Pgt SP)` ltbRgt Similarly, x = a (for S'P lt SP).
Therefore, the locus is `x^(2)=a^(2)`

Again, let (h, k) be the excenter opposite `angleS'`. Then ,
`h=(2a^(2)e sec theta+a^(2)e^(2)sectheta+a^(2)e^(2) sec theta-a^(2)e)/(2a+2ae)`
`aesec theta`
`"and "k=(2aeb tan theta)/(2a+2ae)`
Therefore, the locus is a hyperbola.