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Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002]
- a)x = ±( y+ 2a)
- b)y = ± ( x+ 2a)
- c)x = ± ( y+a)
- d)y = ± ( x+a)
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax ...
Any tangent to the parabola y2 = 8ax is
...(i)
...(i)If (i) is a tangent to the circle, x2 + y2 = 2a2 then,
⇒ m2(1 + m2) = 2 ⇒ (m2 + 2)(m2 – 1) = 0 ⇒ m = ± 1.
So from (i), y = ± (x + 2a).
So from (i), y = ± (x + 2a).
Most Upvoted Answer
Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax ...
The equation of the circle is x^2 + y^2 = 2a^2.
The equation of the parabola is y^2 = 8ax.
To find the common tangents, we can use the fact that the tangents to a circle and a parabola intersect at the point of tangency.
Let's find the point of tangency first. Since the parabola is symmetric with respect to the y-axis, the x-coordinate of the point of tangency is 0. Let's denote the y-coordinate as h.
Substituting x = 0 into the equation of the parabola, we get h^2 = 0, which implies h = 0. Therefore, the point of tangency is (0, 0).
Now, let's find the equations of the tangents.
For the circle, the equation of a tangent at point (x1, y1) is given by (x - x1)(x1) + (y - y1)(y1) = 2a^2.
Substituting (x1, y1) = (0, 0), we get x(0) + y(0) = 2a^2, which simplifies to x = 2a^2.
For the parabola, the equation of a tangent at point (x2, y2) is given by (x - x2)(x2) + (y - y2)(y2) = 8a(x2).
Substituting (x2, y2) = (0, 0), we get x(0) + y(0) = 0, which simplifies to x = 0.
Therefore, the two common tangents are x = 2a^2 and x = 0.
The equation of the parabola is y^2 = 8ax.
To find the common tangents, we can use the fact that the tangents to a circle and a parabola intersect at the point of tangency.
Let's find the point of tangency first. Since the parabola is symmetric with respect to the y-axis, the x-coordinate of the point of tangency is 0. Let's denote the y-coordinate as h.
Substituting x = 0 into the equation of the parabola, we get h^2 = 0, which implies h = 0. Therefore, the point of tangency is (0, 0).
Now, let's find the equations of the tangents.
For the circle, the equation of a tangent at point (x1, y1) is given by (x - x1)(x1) + (y - y1)(y1) = 2a^2.
Substituting (x1, y1) = (0, 0), we get x(0) + y(0) = 2a^2, which simplifies to x = 2a^2.
For the parabola, the equation of a tangent at point (x2, y2) is given by (x - x2)(x2) + (y - y2)(y2) = 8a(x2).
Substituting (x2, y2) = (0, 0), we get x(0) + y(0) = 0, which simplifies to x = 0.
Therefore, the two common tangents are x = 2a^2 and x = 0.
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Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002]a)x = ±( y+ 2a)b)y = ± ( x+ 2a)c)x = ± ( y+a)d)y = ± ( x+a)Correct answer is option 'B'. Can you explain this answer?
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