The angle between a pair of tangents drawn from a point P to the hyperbola `y^2 = 4ax` is `45^@`. Show that the locus of the point P is hyperbola.

Explanation

Let P `(alpha, beta)` be any point on the locus. Equation of pari of tangents from `P (alpha, beta)` to the parabola `y^(2) = 4 ax` is
`[beta y - 2a (x + alpha)]^(2) = (beta^(2 - 4 alpha alpha) (y^(2) - 4 ax)`
`[:' T^(2) = S.S_(1)]`
`= beta^(2) y^(2) + 4 alpha^(2) (x^(2) + alpha^(2) + 2x. alpha) - 4alpha beta y (x + alpha)`
`= beta^(2) y^(2) - 4 beta^(2) ax - 4 alpha alpha y^(2) + 16 a^(2) alpha x`
`implies beta^(2) y^(2) + 4 a^(2) x^(2) + 4a^(2) alpha^(2) + 8x alpha a^(2)`
`beta^(2) y^(2) - 4 beta^(2) ax - 4 alpha alpha y^(2) + 16 a^(2) alpha x - 4a beta x y - 4 alpha beta alpha y`.....(i)
Now, coefficeint of `x^(2) = 4 a^(2)`
coefficient of xy = - `4 a beta`
coefficeint of `y^(2) = 4 a alpha`
Again angle between the two of Eqs. (i) is given as `45^(@)`
`:. tan 45^(@) = (2 sqrt(h^(2) - ab))/(a + b)`
`implies 1 = (2 sqrt(h^(2) - ab))/(a + b)`
`implies a + b = 2 sqrt(h^(2) - ab)`
`implies (4a^(2) + 4 a alpha^(2))^(2) = 4 [4 a^(2) beta^(2) - (4a^(2)) (4 a alpha)]`
`16 a^(2) (a + alpha)^(2) = 4 4a^(2) [beta^(2) - 4 a alpha]`
`implies alpha^(2) + 6 a alpha + a^(2) - beta^(2) = 0`
`implies (alpha + 3 a )^(2) - beta^(2) = 8 a^(2)`