1.The normal to the curve x2+2xy−3y2=0x^2 + 2xy - 3y^2 = 0x2+2xy−3y2=0, at (1,1)(1, 1)(1,1):a.meets the curve again in the third quadrantb.meets the curve again in the fourth quadrantc.does not meet the curve againd.meets the curve again in the second quadrantLogin to continueOnly logged in users canattempt or see the solution.
