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Let S be the focus of the parabola y2 = 8x and PQ be the common chord of the circle x2 + y2 – 2x – 4y = 0 and the given parabola. The area of DPQS is
- a)4 sq units
- b)3 sq units
- c)2 sq units
- d)8 sq units
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Let S be the focus of the parabola y2 = 8x and PQ be the common chord ...
The parametr ic equation s of the parabola y2 = 8x are x = 2t2 and y = 4t.
and the given equation of circle is x2 + y2 – 2x – 4y = 0
On putting x = 2t2 and y = 4t in circle we get
4t4 + 16t2 – 4t2 – 16t = 0
⇒ 4t2 + 12t2 – 16t = 0
⇒ 4t (t3 + 3t – 4) = 0
⇒ t(t – 1) (t2 + t + 4) = 0
⇒ t = 0, t = 1
Thus the coordinates of points of intersection of the circle and the parabola are Q (0, 0) and P(2, 4). Clearly these are diametrically opposite points on the circle.
The coordinates of the focus S of the parabola are (2, 0) which lies on the circle.
= 4 sq. units.
and the given equation of circle is x2 + y2 – 2x – 4y = 0
On putting x = 2t2 and y = 4t in circle we get
4t4 + 16t2 – 4t2 – 16t = 0
⇒ 4t2 + 12t2 – 16t = 0
⇒ 4t (t3 + 3t – 4) = 0
⇒ t(t – 1) (t2 + t + 4) = 0
⇒ t = 0, t = 1
Thus the coordinates of points of intersection of the circle and the parabola are Q (0, 0) and P(2, 4). Clearly these are diametrically opposite points on the circle.
The coordinates of the focus S of the parabola are (2, 0) which lies on the circle.
= 4 sq. units.Free Test
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Community Answer
Let S be the focus of the parabola y2 = 8x and PQ be the common chord ...
The equation of the parabola is y^2 = 8x. This is a standard form equation of a parabola with the vertex at the origin and the focus at (1,0).
The equation of the circle is x^2 + y^2 = r^2, where r is the radius of the circle.
To find the common chord of the circle and the parabola, we need to find the points where the parabola and the circle intersect.
Substituting y^2 = 8x into the equation of the circle, we get x^2 + 8x = r^2.
Completing the square, we have (x + 4)^2 = r^2 + 16.
Taking the square root of both sides, we get x + 4 = ±√(r^2 + 16).
Simplifying, we have x = -4 ± √(r^2 + 16).
Substituting this value of x into the equation of the parabola, we get (y^2)^2 = 8(-4 ± √(r^2 + 16)).
Simplifying further, we have y^4 = -32y^2 ± 64y√(r^2 + 16) + 256.
This is a quartic equation in y. To find the common chord, we need to solve this equation for y and find the values of y that satisfy both the parabola and the circle.
However, it is not possible to find the exact values of y that satisfy both equations without knowing the value of r.
Therefore, the common chord of the circle and the parabola cannot be determined without additional information.
The equation of the circle is x^2 + y^2 = r^2, where r is the radius of the circle.
To find the common chord of the circle and the parabola, we need to find the points where the parabola and the circle intersect.
Substituting y^2 = 8x into the equation of the circle, we get x^2 + 8x = r^2.
Completing the square, we have (x + 4)^2 = r^2 + 16.
Taking the square root of both sides, we get x + 4 = ±√(r^2 + 16).
Simplifying, we have x = -4 ± √(r^2 + 16).
Substituting this value of x into the equation of the parabola, we get (y^2)^2 = 8(-4 ± √(r^2 + 16)).
Simplifying further, we have y^4 = -32y^2 ± 64y√(r^2 + 16) + 256.
This is a quartic equation in y. To find the common chord, we need to solve this equation for y and find the values of y that satisfy both the parabola and the circle.
However, it is not possible to find the exact values of y that satisfy both equations without knowing the value of r.
Therefore, the common chord of the circle and the parabola cannot be determined without additional information.
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Let S be the focus of the parabola y2 = 8x and PQ be the common chord of the circle x2 + y2 – 2x – 4y = 0 and the given parabola. The area of DPQS isa)4 sq unitsb)3 sq unitsc)2 sq unitsd)8 sq unitsCorrect answer is option 'A'. Can you explain this answer?
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