if the slope of the tangent to the curve x^2 -2xy + 4y=0 at a point it is -3/2 then find the equations of the tangent and the normal at that point
Answers
Given : slope of the tangent to the curve x^2 -2xy + 4y=0 at a point it is -3/2
To Find : the equations of the tangent and the normal at that point
Solution:
x²-2xy + 4y=0
=> 2x - 2xdy/dx - 2y + 4dy/dx = 0
=> x - xdy/dx - y + 2dy/dx = 0
=> dy/dx(2 - x) = y - x
=> dy/dx = (y - x)/(2 - x)
(y - x)/(2 - x) = -3/2
=> 2y - 2x = -6 + 3x
=>2y = 5x - 6
x²-2xy + 4y=0
=> x²- x(2y) + 2(2y)=0
=> x² - x( 5x - 6) + 2(5x - 6) =0
=> x² - 5x² + 6x + 10x - 12 = 0
=> -4x² + 16x - 12 = 0
=> x² - 4x + 3 = 0=
=> (x - 3)(x - 1) = 0
=> x = 3 , 1
2y = 5x - 6
=> 2y = 9 or 2y = -1
=> y = 9/2 , y = - 1/2
point ( 3 , 9/2) and ( 1 , - 1/2)
Slope = - 3/2
Tangent Equation at 3 , 9/2
=> y - 9/2 = (-3/2)(x - 3)
=> 2y - 9 = -3x + 9
=> 3x + 2y = 18
Normal slope = 2/3
Equation of Normal at Tangent Equation at 3 , 9/2
=> y - 9/2 = (2/3)(x - 3)
=> 6y - 27 = 4x - 12
=> 6y = 4x + 15
Similarly at ( 1 , - 1/2)
Tangent
3x + 2y = 2
Normal
6y = 4x - 7
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Answer:
x²-2xy + 4y=0
=> x²- x(2y) + 2(2y)=0
=> x² - x( 5x - 6) + 2(5x - 6) =0
=> x² - 5x² + 6x + 10x - 12 = 0
=> -4x² + 16x - 12 = 0
=> x² - 4x + 3 = 0=
=> (x - 3)(x - 1) = 0
=> x = 3 , 1
Step-by-step explanation: