1.The transverse axis of a hyperbola is along the x-axis and its length is 2a2a2a. The vertex bisects the segment joining the centre and the focus. The equation of the hyperbola isa.6x2−y2=3a26x^2 - y^2 = 3a^26x2−y2=3a2b.x2−3y2=3a2x^2 - 3y^2 = 3a^2x2−3y2=3a2c.x2−6y2=3a2x^2 - 6y^2 = 3a^2x2−6y2=3a2d.3x2−y2=3a23x^2 - y^2 = 3a^23x2−y2=3a2Login to continueOnly logged in users canattempt or see the solution.