Application of DerivativeHard
Question
The normal to the curve, x2 + 2xy - 3y2 = 0, at (1, 1)
Options
A.meets the curve again in the third quadrant
B.meets the curve again in the fourth quadrant
C.does not meet the curve again
D.meets the curve again in the second quadrant
Solution
x2 + 3xy - xy - 3y2 = 0
x(x + 3y) - y(x + 3y) = 0
(x - y) (x + 3y) = 0
Normal at (1, 1) will be x + y = 2
Now, x + y = 2
x + 3y = 0
2y = - 2, y = -1
& x = 3
∴ (3, - 1)
which is in fourth quadrant.
x(x + 3y) - y(x + 3y) = 0
(x - y) (x + 3y) = 0
Normal at (1, 1) will be x + y = 2
Now, x + y = 2
x + 3y = 0
2y = - 2, y = -1
& x = 3
∴ (3, - 1)
which is in fourth quadrant.
Create a free account to view solution
View Solution FreeMore Application of Derivative Questions
The tangent to the curve x = a(θ − sinθ ), y = a(1 + cosθ) at the points = (2k + 1)π, k∈...Which of the following pair(s) of curves is/are orthogonal....The radius of an air bubble is increasing at the rate of 0.5 cm/sec. The rate by which the volume of the bubble is incre...The tangent to the curve 3xy2 - 2x2y = 1 at (1,1) meets the curve again at the point -...The points on the curve 9y2 = x3 where the normal to the curve cuts equal intercepts from the axes are-...