The equation `2x^2+4xy-py^2+4x+qy+1=0` will represent two mutually perpendicular straight lines , if

Explanation

The given equation represents a pair of lines perpendicular to each other.
`therefore` (coefficient of `x^(2)`) + (coefficient of `y^(2)`) = 0
`therefore 2-p = 0 " " therefore p = 2`
Comparing this equation with
`ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c = 0`, we get,
`a = 2, h = 2, b = -2, f = (q)/(2) and c = 1`.
`therefore D =|{:(a,,h,,g),(h,,b,,f),(g,,f,,c):}|=|{:(2,,2,,2),(2,,-2,,(q)/(2)),(2,,(q)/(2),,1):}|`
` =2(-2-(q^(2))/(4))-2(2-q)+2(q+4)`
` = -4-(q^(2))/(2) -4 + 2q + 2q +8`
` = -(q^(2))/(2) + 4q`
SInce the given represents a pair of lines, `D = 0`.
`therefore -(q^(2))/(2) +4q =0 " "therefore q^(2) -8q = 0`
`therefore q(q-8)=0" "therefore q = 0 orq = 8`
Hence, p = 2 and q = 0 or 8.