Choose the correct option from the given alternatives : The normal to the curve x2 + 2xy – 3y2 = 0 at (1, 1)

Question

KollegeApply · Accepted Answer

meets the curve again in fourth quadrant Explanation: x2 + 2xy - 3y2 = 0 ∴ `2x + 2(xdy/dx + y xx 1) - 3 xx 2ydy/dx = 0 ∴ (2x - 6y)dy/dx = -2x - 2y ∴ dy/dx = (-(x + y))/(x - 3y) ∴ (dy/dx)_("at" (1, 1)) = (-1(1 + 1))/(1 - 3)` = 1 = slope of the tangent at (1, 1) = slope of the normal at (1, 1) is – 1 = equation of the normal is y – 1 = – 1(x – 1) = – x + 1 ∴ x + y = 2 ∴ y = 2 – x Substituting y = 2 – x in x2 + 2xy – 3y2 = 0, we get x2 + 2x(2 – x) – 3(2 – x)2 = 0 ∴ x2 + 4x – 2x2 – 3(4 – 4x + x2) = 0 ∴ x2 – 4x + 3 = 0 ∴ (x – 1)(x – 3) = 0 ∴ x = 1, x = 3 When x = 1, y = 2 – 1 = 1 When x = 3, y = 2 – 3 = –1 ∴ The normal at (1, 1) meets the curve at (3, –1) which is in the fourth quadrant.

AChoose Your Answer