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Let PQ be a chord of the parabola y2 = 4x. A circle drawn with PQ as diameter passes through the vertex V of the parabola. If area of ΔPVQ = 20 square units then coordinates of P are
- a)(-16, -8)
- b)(-16, 8)
- c)(16, -8)
- d)(16, 8)
Correct answer is option 'C,D'. Can you explain this answer?
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Most Upvoted Answer
Let PQ be a chord of the parabola y2 = 4x. A circle drawn with PQ as ...
Slope of
Slope of
Equation of VQ 
Solving it with y2 = 4x
x(t2x - 16) = 0
x = 0, x = 16/t2
Area of ΔPVQ = 
PV2. VQ2 = 1600
t2 + 4 = ± 5t
t = ± 1, ± 4
Hence, P (16 ± 8), (1, ±2)
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Community Answer
Let PQ be a chord of the parabola y2 = 4x. A circle drawn with PQ as ...
To find the coordinates of point P, we need to understand the properties of the parabola and the circle passing through the vertex.
Properties of the Parabola:
1. The vertex of the parabola y^2 = 4x is (0,0). This is because the equation of the parabola is in the form (y-k)^2 = 4a(x-h), where (h,k) is the vertex.
2. The focus of the parabola is at (a,0) where a is the distance between the vertex and the focus. In this case, a = 1.
Properties of the Circle:
1. The diameter of the circle passes through the vertex of the parabola, which is (0,0).
2. The center of the circle lies on the perpendicular bisector of the chord PQ.
Using these properties, we can find the coordinates of point P.
1. Let the coordinates of point P be (x, y).
2. Since PQ is a chord of the parabola, the coordinates of Q are (-x, y). This is because the parabola is symmetric with respect to the y-axis.
3. Since the circle passes through the vertex V (0,0), the center of the circle lies on the perpendicular bisector of PQ.
4. The midpoint of PQ is ((x + (-x))/2, (y + y)/2) = (0, y).
5. The slope of PQ is (y - y)/(-x - x) = 0.
6. The equation of the perpendicular bisector of PQ is y = y.
7. Since the center of the circle lies on the perpendicular bisector, the x-coordinate of the center is 0.
8. The y-coordinate of the center is the same as the y-coordinate of the midpoint, which is y.
9. The equation of the circle is therefore x^2 + (y - y)^2 = r^2, where r is the radius of the circle.
10. Simplifying the equation, we get x^2 + y^2 = r^2.
Since the circle passes through P, the coordinates (x, y) must satisfy the equation of the circle. Substituting the coordinates of P into the equation, we get x^2 + y^2 = r^2.
Given that the area of triangle PVQ is 20 square units, we can use the formula for the area of a triangle to find the length of PQ.
1. The length of PQ is given by the formula: area = (1/2) * base * height.
2. In this case, the base is the distance between the x-coordinates of P and Q, which is x - (-x) = 2x.
3. The height is the distance between the y-coordinate of P and the y-coordinate of Q, which is y - y = 0.
4. Substituting the values into the formula, we get 20 = (1/2) * 2x * 0.
5. Simplifying, we get 20 = 0.
6. This is not possible, so there must be an error in the question or the given information.
Therefore, the answer cannot be determined and none of the options provided are correct.
Properties of the Parabola:
1. The vertex of the parabola y^2 = 4x is (0,0). This is because the equation of the parabola is in the form (y-k)^2 = 4a(x-h), where (h,k) is the vertex.
2. The focus of the parabola is at (a,0) where a is the distance between the vertex and the focus. In this case, a = 1.
Properties of the Circle:
1. The diameter of the circle passes through the vertex of the parabola, which is (0,0).
2. The center of the circle lies on the perpendicular bisector of the chord PQ.
Using these properties, we can find the coordinates of point P.
1. Let the coordinates of point P be (x, y).
2. Since PQ is a chord of the parabola, the coordinates of Q are (-x, y). This is because the parabola is symmetric with respect to the y-axis.
3. Since the circle passes through the vertex V (0,0), the center of the circle lies on the perpendicular bisector of PQ.
4. The midpoint of PQ is ((x + (-x))/2, (y + y)/2) = (0, y).
5. The slope of PQ is (y - y)/(-x - x) = 0.
6. The equation of the perpendicular bisector of PQ is y = y.
7. Since the center of the circle lies on the perpendicular bisector, the x-coordinate of the center is 0.
8. The y-coordinate of the center is the same as the y-coordinate of the midpoint, which is y.
9. The equation of the circle is therefore x^2 + (y - y)^2 = r^2, where r is the radius of the circle.
10. Simplifying the equation, we get x^2 + y^2 = r^2.
Since the circle passes through P, the coordinates (x, y) must satisfy the equation of the circle. Substituting the coordinates of P into the equation, we get x^2 + y^2 = r^2.
Given that the area of triangle PVQ is 20 square units, we can use the formula for the area of a triangle to find the length of PQ.
1. The length of PQ is given by the formula: area = (1/2) * base * height.
2. In this case, the base is the distance between the x-coordinates of P and Q, which is x - (-x) = 2x.
3. The height is the distance between the y-coordinate of P and the y-coordinate of Q, which is y - y = 0.
4. Substituting the values into the formula, we get 20 = (1/2) * 2x * 0.
5. Simplifying, we get 20 = 0.
6. This is not possible, so there must be an error in the question or the given information.
Therefore, the answer cannot be determined and none of the options provided are correct.
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Let PQ be a chord of the parabola y2 = 4x. A circle drawn with PQ as diameter passes through the vertex V of the parabola. If area of ΔPVQ = 20 square units then coordinates of P area)(-16, -8)b)(-16, 8)c)(16, -8)d)(16, 8)Correct answer is option 'C,D'. Can you explain this answer?
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