1.The equation of the normal to the curve 3x2−y2=83x^2 - y^2 = 83x2−y2=8, which is parallel to the line x+3y=10x + 3y = 10x+3y=10, isa.x+3y+6=0x + 3y + 6 = 0x+3y+6=0b.x+3y−3=0x + 3y - 3 = 0x+3y−3=0c.x+3y+8=0x + 3y + 8 = 0x+3y+8=0d.x+3y−4=0x + 3y - 4 = 0x+3y−4=0Login to continueOnly logged in users canattempt or see the solution.
