Correct Answer - B
Let the equation of the hyperbola be
`2x^(2)+2y^(2)+5xy+lambda=0`
It passes through (1,1). Therefore,
`2+2+5+lambda=0`
`"or "lambda=-9`
So, the hyperbola is
`2x^(2)+2y^(2)+5xy=9`
The equation of the tangent at `(-1,7//2)` is given by
`2x(-1)+2y((7)/(2))+5(x(7//2)+(-1)y)/(2)=9`
`"or "3x+2y=4`