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If the parametric equations of the parabola are given by x = 4t2 - 2t + 1; y = 3t2 + t + 1 and the vertex of the parabola also satisfies y - x = k/100, then the area of the circle x2 + y2 + 12x -10y + 2k = 0 in square units is
- a)π
- b)3π
- c)4π
- d)5π
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If the parametric equations of the parabola are given by x = 4t2 - 2t ...
Eliminating t, we get (3x - 4y + 2)2 = 16x + 12y - 27. Vertex is 
Hence k = 29 and the area is 3π.
Hence k = 29 and the area is 3π.
Most Upvoted Answer
If the parametric equations of the parabola are given by x = 4t2 - 2t ...
To find the vertex of the parabola, we can set the derivative of y with respect to t equal to 0:
dy/dt = 6t - 1 = 0
6t = 1
t = 1/6
Substituting t = 1/6 into the parametric equations, we get:
x = 4(1/6)^2 - 2(1/6) + 1 = 1/9
y = 3(1/6)^2 - 1/6 + 1 = 1/18
So the vertex of the parabola is (1/9, 1/18).
Now, let's find the equation of the circle. We have:
x^2 + y^2 - 12x - 10y + 2k = 0
Completing the square for x and y terms, we get:
(x^2 - 12x + 36) + (y^2 - 10y + 25) - 36 - 25 + 2k = 0
(x - 6)^2 + (y - 5)^2 + 2k - 61 = 0
Comparing this with the standard equation of a circle, we can see that the center of the circle is (6, 5) and the radius is sqrt(61 - 2k).
To find the area of the circle, we can use the formula:
Area = π * r^2
Area = π * (sqrt(61 - 2k))^2
Area = π * (61 - 2k)
So the area of the circle is π(61 - 2k) square units.
dy/dt = 6t - 1 = 0
6t = 1
t = 1/6
Substituting t = 1/6 into the parametric equations, we get:
x = 4(1/6)^2 - 2(1/6) + 1 = 1/9
y = 3(1/6)^2 - 1/6 + 1 = 1/18
So the vertex of the parabola is (1/9, 1/18).
Now, let's find the equation of the circle. We have:
x^2 + y^2 - 12x - 10y + 2k = 0
Completing the square for x and y terms, we get:
(x^2 - 12x + 36) + (y^2 - 10y + 25) - 36 - 25 + 2k = 0
(x - 6)^2 + (y - 5)^2 + 2k - 61 = 0
Comparing this with the standard equation of a circle, we can see that the center of the circle is (6, 5) and the radius is sqrt(61 - 2k).
To find the area of the circle, we can use the formula:
Area = π * r^2
Area = π * (sqrt(61 - 2k))^2
Area = π * (61 - 2k)
So the area of the circle is π(61 - 2k) square units.
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If the parametric equations of the parabola are given by x = 4t2 - 2t + 1; y = 3t2 + t + 1 and the vertex of the parabola also satisfiesy - x = k/100,then the area of the circlex2 + y2 + 12x -10y + 2k = 0 in square units isa)πb)3πc)4πd)5πCorrect answer is option 'B'. Can you explain this answer?
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