Q. 387. Co-ordinate geometry, Contact points of vertical tangents to a curve

Question 387.
Find the coordinates of all points on the curve (x+y)² + 2(x-y)² =24
at which the tangent is parallel to the y-axis.

Answer 387.
(x + y)^2 + 2(x - y)^2 = 24
=> 3x^2 + 3y^2 - 2xy = 24
=> 6x + 6y dy/dx - 2 (y + x dy/dx) = 0
=> 2 (3x - y) + 2 (3y - x) dy/dx = 0
=> dy/dx = (3x - y)/(x - 3y)

For tangent to be parallel to y-axis, dy/dx is not defined
=> x - 3y = 0
 For point P (x1, y1),
x1 - 3y1 = 0 ... ( 1 )
Also, 3x1^2 + 3y1^2 - 2x1y1 = 24 ... ( 2 )

Plugging x1 = 3y1 from ( 1 ) in ( 2 ),
3 (3y1)^2 + 3y1^2 - 2(3y1) * y1 = 24
=> 24y1^2 = 24
=> y1 = -1 or +1
=> x1 = 3y1 = - 3 or +3
=> points are (-3, -1) and (3, 1).
 
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