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From the focus of the parabola y2 = 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is
- a)(x – 2)2 + y2 = 3
- b)(x – 2)2 + y2 = 9
- c)(x + 2)2 + y2 = 9
- d)None
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
From the focus of the parabola y2= 8x as centre, a circle is described...
Focus of parabola y2 = 8x is (2,0). Equation of circle with centre (2,0) is (x−2)2 + y2 = r2
Let AB is common chord and Q is mid point i.e. (1,0)
AQ2 = y2 = 8x
= 8×1 = 8
∴ r2 = AQ2 + QS2
= 8 + 1 = 9
So required circle is (x−2)2 + y2 = 9
Let AB is common chord and Q is mid point i.e. (1,0)
AQ2 = y2 = 8x
= 8×1 = 8
∴ r2 = AQ2 + QS2
= 8 + 1 = 9
So required circle is (x−2)2 + y2 = 9
Most Upvoted Answer
From the focus of the parabola y2= 8x as centre, a circle is described...
The vertex of the parabola y^2 = 8x is (0,0) and the focus is (2,0). Let the center of the circle be (h,k) and its radius be r. Then, the equation of the circle is (x-h)^2 + (y-k)^2 = r^2.
Since the common chord of the parabola and circle is equidistant from the vertex and focus, it must be perpendicular to the axis of the parabola, which is the x-axis. Let the common chord intersect the x-axis at (a,0). Then, the distance from (a,0) to the vertex (0,0) is a, and the distance from (a,0) to the focus (2,0) is a-2.
By definition, the distance from a point on the circle to the focus is equal to the distance from the same point on the circle to the common chord. Therefore, we have:
√[(h-a)^2 + k^2] = √[(h-a+2)^2 + k^2-a^2]
Simplifying this equation, we get:
(h-a)^2 = (h-a+2)^2 + (a^2-k^2)
h^2 - 2ah + a^2 = h^2 - 4ah + 4a + a^2 - k^2
2ah = 4a - k^2
h = (2a - k^2) / (2a)
Since the common chord is perpendicular to the x-axis, the line passing through (a,0) and (h,k) has a slope of (k-0)/(h-a) = k / ((2a-k^2)/(2a)-a) = -k / (k^2-2a^2). This line must also be perpendicular to the tangent line of the parabola at (a^2/8,a). The slope of the tangent line at (a^2/8,a) is a/2, so the slope of the line passing through (a,0) and (h,k) is -2/a.
Therefore, we have:
-k / (k^2-2a^2) = -2/a
k^2 - 2a^2 = -2ak
Solving for k, we get:
k = ±√(2a^3 - a^2)
Substituting h and k into the equation of the circle, we get:
(x - (2a-k^2)/(2a))^2 + (y ± √(2a^3 - a^2))^2 = (2a^3 - a^2) / (2a)
Simplifying this equation, we get:
x^2 + y^2 - (4a-k^2)x ± 2√(2a^3 - a^2)y + 4a^2 = 0
Therefore, the equation of the circle is:
(x^2 + y^2 - (4a-k^2)x + 4a^2)^2 + 8(2a^3 - a^2)y^2 = (2a)(2a^3 - a^2)^2
Answer: (c) (x^2 + y^2
Since the common chord of the parabola and circle is equidistant from the vertex and focus, it must be perpendicular to the axis of the parabola, which is the x-axis. Let the common chord intersect the x-axis at (a,0). Then, the distance from (a,0) to the vertex (0,0) is a, and the distance from (a,0) to the focus (2,0) is a-2.
By definition, the distance from a point on the circle to the focus is equal to the distance from the same point on the circle to the common chord. Therefore, we have:
√[(h-a)^2 + k^2] = √[(h-a+2)^2 + k^2-a^2]
Simplifying this equation, we get:
(h-a)^2 = (h-a+2)^2 + (a^2-k^2)
h^2 - 2ah + a^2 = h^2 - 4ah + 4a + a^2 - k^2
2ah = 4a - k^2
h = (2a - k^2) / (2a)
Since the common chord is perpendicular to the x-axis, the line passing through (a,0) and (h,k) has a slope of (k-0)/(h-a) = k / ((2a-k^2)/(2a)-a) = -k / (k^2-2a^2). This line must also be perpendicular to the tangent line of the parabola at (a^2/8,a). The slope of the tangent line at (a^2/8,a) is a/2, so the slope of the line passing through (a,0) and (h,k) is -2/a.
Therefore, we have:
-k / (k^2-2a^2) = -2/a
k^2 - 2a^2 = -2ak
Solving for k, we get:
k = ±√(2a^3 - a^2)
Substituting h and k into the equation of the circle, we get:
(x - (2a-k^2)/(2a))^2 + (y ± √(2a^3 - a^2))^2 = (2a^3 - a^2) / (2a)
Simplifying this equation, we get:
x^2 + y^2 - (4a-k^2)x ± 2√(2a^3 - a^2)y + 4a^2 = 0
Therefore, the equation of the circle is:
(x^2 + y^2 - (4a-k^2)x + 4a^2)^2 + 8(2a^3 - a^2)y^2 = (2a)(2a^3 - a^2)^2
Answer: (c) (x^2 + y^2
Community Answer
From the focus of the parabola y2= 8x as centre, a circle is described...
(X-2)2=x2=2(2)=4×2=8Y2=1X=8+Y=1 =9
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From the focus of the parabola y2= 8x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle isa)(x –2)2+ y2= 3b)(x –2)2+ y2= 9c)(x + 2)2+ y2= 9d)NoneCorrect answer is option 'B'. Can you explain this answer?
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