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The area enclosed by the parabola y2 = 2x and its tangents through the point (-2 , 0) is
- a)8/3
- b)4
- c)3
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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The area enclosed by the parabolay2 = 2x and its tangents through the ...
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The area enclosed by the parabolay2 = 2x and its tangents through the ...
Given parabola is y^2 = 2x.
Tangents at the point (-2,0) will be y = 0 and y = mx + 2m + 2, where m is the slope of the tangent.
Let P be the point of intersection of the tangents and Q be the point where the tangent intersects the x-axis.
Let Q1 and Q2 be the x-intercepts of the tangents.
Using the equation of the tangents, we get
Q = (-2/m, 0)
Q1 = (-2 - 2m - 2/m, 0)
Q2 = (-2 + 2m - 2/m, 0)
The coordinates of P can be found by equating the slopes of PQ1 and PQ2 as they are perpendicular to each other.
Slope of PQ1 = (0 - 0)/(P.x - Q1.x) = -1/(P.x + 2m + 2)
Slope of PQ2 = (0 - 0)/(P.x - Q2.x) = -1/(P.x - 2m + 2)
Equating the slopes, we get
P.x^2 - 4m^2 + 8m - 4 = 0
Solving for m, we get
m = (2 - P.x)/2
Substituting this value of m in the equation of the tangents, we get
y = mx + 2m + 2 = (P.x - 2)x + P.x + 2
The x-coordinate of P can be found by equating the y-coordinates of P with the equation of the parabola y^2 = 2x.
Substituting y = (P.x - 2)x + P.x + 2 in y^2 = 2x, we get
(P.x - 2)^2 x^2 + 2(P.x - 2)(P.x + 2)x + (P.x + 2)^2 - 8P.x = 0
Solving for P.x, we get
P.x = 2/3
The area enclosed by the parabola and the tangents is given by
A = 2∫(0,2/3) y dx
Substituting y = (P.x - 2)x + P.x + 2, we get
A = 2∫(0,2/3) [(P.x - 2)x + P.x + 2] dx
A = 8/3
Hence, the correct option is A.
Tangents at the point (-2,0) will be y = 0 and y = mx + 2m + 2, where m is the slope of the tangent.
Let P be the point of intersection of the tangents and Q be the point where the tangent intersects the x-axis.
Let Q1 and Q2 be the x-intercepts of the tangents.
Using the equation of the tangents, we get
Q = (-2/m, 0)
Q1 = (-2 - 2m - 2/m, 0)
Q2 = (-2 + 2m - 2/m, 0)
The coordinates of P can be found by equating the slopes of PQ1 and PQ2 as they are perpendicular to each other.
Slope of PQ1 = (0 - 0)/(P.x - Q1.x) = -1/(P.x + 2m + 2)
Slope of PQ2 = (0 - 0)/(P.x - Q2.x) = -1/(P.x - 2m + 2)
Equating the slopes, we get
P.x^2 - 4m^2 + 8m - 4 = 0
Solving for m, we get
m = (2 - P.x)/2
Substituting this value of m in the equation of the tangents, we get
y = mx + 2m + 2 = (P.x - 2)x + P.x + 2
The x-coordinate of P can be found by equating the y-coordinates of P with the equation of the parabola y^2 = 2x.
Substituting y = (P.x - 2)x + P.x + 2 in y^2 = 2x, we get
(P.x - 2)^2 x^2 + 2(P.x - 2)(P.x + 2)x + (P.x + 2)^2 - 8P.x = 0
Solving for P.x, we get
P.x = 2/3
The area enclosed by the parabola and the tangents is given by
A = 2∫(0,2/3) y dx
Substituting y = (P.x - 2)x + P.x + 2, we get
A = 2∫(0,2/3) [(P.x - 2)x + P.x + 2] dx
A = 8/3
Hence, the correct option is A.
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The area enclosed by the parabolay2 = 2x and its tangents through the point (-2 , 0) isa)8/3b)4c)3d)none of theseCorrect answer is option 'A'. Can you explain this answer?
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