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Equation of normal to the curve y = x(2 - x) at the point (2, 0) is
- a)x - 2y = 2
- b)2x + y = 4
- c)x - 2y + 2 = 0
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Equation of normal to the curve y = x(2 - x) at the point (2, 0) isa)x...
Solution:
Given curve is y = x(2 - x)
Let P(x, y) be the point on the curve where x = 2.
Then, y = 2(2 - 2) = 0.
So, the point P is (2, 0).
The slope of the tangent to the curve at P is given by the derivative of y w.r.t. x at x = 2.
dy/dx = (2 - 2x)
So, the slope of the tangent at P is (2 - 2(2)) = -2.
The slope of the normal to the curve at P is the negative reciprocal of the slope of the tangent at P.
So, the slope of the normal at P is 1/2.
Using the point-slope form of the equation of a line, the equation of the normal at P is given by:
y - 0 = (1/2)(x - 2)
Simplifying, we get:
x - 2y = 2
Hence, the correct option is (a) x - 2y = 2.
Given curve is y = x(2 - x)
Let P(x, y) be the point on the curve where x = 2.
Then, y = 2(2 - 2) = 0.
So, the point P is (2, 0).
The slope of the tangent to the curve at P is given by the derivative of y w.r.t. x at x = 2.
dy/dx = (2 - 2x)
So, the slope of the tangent at P is (2 - 2(2)) = -2.
The slope of the normal to the curve at P is the negative reciprocal of the slope of the tangent at P.
So, the slope of the normal at P is 1/2.
Using the point-slope form of the equation of a line, the equation of the normal at P is given by:
y - 0 = (1/2)(x - 2)
Simplifying, we get:
x - 2y = 2
Hence, the correct option is (a) x - 2y = 2.
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Community Answer
Equation of normal to the curve y = x(2 - x) at the point (2, 0) isa)x...
Solution:
Given, y = x(2 - x)
Differentiating both sides w.r.t x, we get
dy/dx = 2x - 1
At the point (2, 0), the slope of the tangent is
dy/dx = 2(2) - 1 = 3
The slope of the normal at this point will be the negative reciprocal of the slope of the tangent.
Therefore, slope of the normal = -1/3
Let the equation of the normal be y = mx + c, where m is the slope of the normal and c is the y-intercept.
Substituting the point (2, 0) in the equation, we get
0 = -2/3 + c
c = 2/3
Therefore, the equation of the normal is y = (-1/3)x + 2/3
Multiplying both sides by 3, we get
3y = -x + 2
Adding x to both sides, we get
x - 3y + 2 = 0
Therefore, the equation of the normal to the curve y = x(2 - x) at the point (2, 0) is x - 3y + 2 = 0, which is equivalent to x - 2y = 2. Hence, option (a) is the correct answer.
Given, y = x(2 - x)
Differentiating both sides w.r.t x, we get
dy/dx = 2x - 1
At the point (2, 0), the slope of the tangent is
dy/dx = 2(2) - 1 = 3
The slope of the normal at this point will be the negative reciprocal of the slope of the tangent.
Therefore, slope of the normal = -1/3
Let the equation of the normal be y = mx + c, where m is the slope of the normal and c is the y-intercept.
Substituting the point (2, 0) in the equation, we get
0 = -2/3 + c
c = 2/3
Therefore, the equation of the normal is y = (-1/3)x + 2/3
Multiplying both sides by 3, we get
3y = -x + 2
Adding x to both sides, we get
x - 3y + 2 = 0
Therefore, the equation of the normal to the curve y = x(2 - x) at the point (2, 0) is x - 3y + 2 = 0, which is equivalent to x - 2y = 2. Hence, option (a) is the correct answer.
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