about to only mathematics

Options

A. a hyperbola
B. a parabola
C. an ellipse
D. a straight line

Explanation

Given, lines are `(1+p)x-py+p(1+p)=0`…….`(i)`
and `(1+p)x-qy+q(1+q)=0`……`(ii)`
On solving Eqs. `(i)` and `(ii)` we get
`C{pq,(1+p)(1+q)}`
`:.` Equation of altitude `CM` passing through `C` and perpendicular to `AB` is
`x=pq`….`(iii)`
` :' "Slope of" (ii) "is" ((1+q)/(q))`
`:.` Slope of altitude `BN` (as shown in figure) is `(-q)/(1+q)`

`:.` Equation of `BN` is `y-0=(-q)/(1+q)(x+p)`
`impliesy=(-q)/((1+q))(x+p)`.......`(iv)`
Let orthocentre of triangle be `H(h,k)`, which is the point of intersection of Eqs. `(iii)` and `(iv)`
On solving Eqs. `(iii)` and `(iv)`, we get
`x=pq` and `y=-pq`
`impliesh=pq` and `k=-pq`
`:. h+k=0`
`:.` Locus of `H(h,k)` is `x+y=0`.