1.Focus (±3,0)(\pm3,0)(±3,0), tangent 2x+y−4=02x+y-4=02x+y−4=0. Equation of hyperbola isa.4x2−5y2=204x^2-5y^2=204x2−5y2=20b.5x2−4y2=205x^2-4y^2=205x2−4y2=20c.4x2−5y2=14x^2-5y^2=14x2−5y2=1d.5x2−4y2=15x^2-4y^2=15x2−4y2=1Login to continueOnly logged in users canattempt or see the solution.
