Factorise the following: `p^2+6p–16`

Answer

`a rarrp, q,r,s; b rarrq, r; c rarrp; d rarrp,s`

Explanation

(a) Obviously, all the points in List II are common to the hyperbola and circle.
(b). The chord of contact of hyperbola w.r.t. (0, -9/4) is
`0(x)-(-(9)/(4))y=9`
`"or "y=4`
Solving this with hyperbola, we have
`x^(2)-16=9`
`"or "x^(2)=25or x= pm5`
Hence, the point of contact are `(pm 5, 4)`.
(c) Obviously, the required point is `(-5, -4)`.
(d) Let the point on the hyperbola be P(h,k) and `Q(-h,k)`. Then
`"Area of triangle"=(1)/(2)|2h||-6-k|=10" (1)"`
Also, points P and Q lie on the hyperbola. Hence,
`h^(2)-k^(2)=9" (2)"`
Clearly, point `(pm5,-4)` satisfy both (1) and (2).