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The equation of a parabola which passes through the intersection of a straight line x + y = 0 and the circle x2 + y2 + 4y = 0 is
- a)y2 = 4x
- b)y2 = x
- c)y2 = 2x
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The equation of a parabola which passes through the intersection of a ...
As the straight line and circle intersect so x = −y satisfies the equation of circle x2 + y2 + 4y = 0
(−y)2 + y2 + 4y = 0 ⇒ 2y2 + 4y = 0 ⇒ 2y(y + 2) = 0 ⇒ y = 0, −2
∴ x = 0, 2
Thus, the interesetion of line and circle is (0,0) and (2, −2).
Since, options are of form y2 = 4ax, so putting (2,−2) in y2 = 4ax
Putting back in standard equation of parabola,
Putting back in standard equation of parabola,
which also satisfies point (0, 0)
Thus, y2 = 2x is the equation of parabola.
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The equation of a parabola which passes through the intersection of a ...
To find the equation of the parabola that passes through the intersection of the given straight line and circle, we need to find the coordinates of the intersection point first.
1. Find the intersection point:
The straight line equation is given as x + y = 0.
Substitute y = -x into the equation of the circle: x^2 + (-x)^2 - 4(-x) = 0.
Simplifying the equation gives: 2x^2 + 4x = 0.
Factor out x: x(2x + 4) = 0.
Solving for x, we get two possible solutions: x = 0 or x = -2.
For x = 0, substitute into the equation of the straight line: 0 + y = 0.
Therefore, the first intersection point is (0, 0).
For x = -2, substitute into the equation of the straight line: -2 + y = 0.
Therefore, the second intersection point is (-2, 2).
2. Find the equation of the parabola:
We have two points that the parabola passes through: (0, 0) and (-2, 2).
Let the equation of the parabola be y^2 = ax.
Substitute the coordinates of the two points into the equation:
For (0, 0): (0)^2 = a(0) => 0 = 0.
For (-2, 2): (2)^2 = a(-2) => 4 = -2a.
From the second equation, we can solve for a: a = -4/2 = -2.
Therefore, the equation of the parabola is y^2 = -2x, which matches option C.
1. Find the intersection point:
The straight line equation is given as x + y = 0.
Substitute y = -x into the equation of the circle: x^2 + (-x)^2 - 4(-x) = 0.
Simplifying the equation gives: 2x^2 + 4x = 0.
Factor out x: x(2x + 4) = 0.
Solving for x, we get two possible solutions: x = 0 or x = -2.
For x = 0, substitute into the equation of the straight line: 0 + y = 0.
Therefore, the first intersection point is (0, 0).
For x = -2, substitute into the equation of the straight line: -2 + y = 0.
Therefore, the second intersection point is (-2, 2).
2. Find the equation of the parabola:
We have two points that the parabola passes through: (0, 0) and (-2, 2).
Let the equation of the parabola be y^2 = ax.
Substitute the coordinates of the two points into the equation:
For (0, 0): (0)^2 = a(0) => 0 = 0.
For (-2, 2): (2)^2 = a(-2) => 4 = -2a.
From the second equation, we can solve for a: a = -4/2 = -2.
Therefore, the equation of the parabola is y^2 = -2x, which matches option C.
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The equation of a parabola which passes through the intersection of a straight linex + y = 0 and the circlex2 + y2 + 4y = 0 isa)y2 = 4xb)y2 = xc)y2 = 2xd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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