1.The curve that passes through the point (2,3)(2, 3)(2,3), and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by:a.(x2)2+(y3)2=2\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 2(2x)2+(3y)2=2b.2y−3x=02y - 3x = 02y−3x=0c.y=6xy = \frac{6}{x}y=x6d.x2+y2=13x^2 + y^2 = 13x2+y2=13Login to continueOnly logged in users canattempt or see the solution.