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Parabola Test - 1

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Parabola Test - 1
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  • Question 1
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    If the line 2x – y + λ = 0 is a diameter of the circle x2+y2+6x−6y+5=0 then λ =

    Solution

    Equation of circle is x2+y2+6x−6y+5=0 

    x2+6x+y2−6y=−5

    Applying completing the square method

    (x+3)2+(y−3)2=(13√)2

    Comparing the above equation with(x−h)2+(y−k)2=(r)2   we get center as (-3,3) and radius as √ 13  .

    As centre of the circlre lies on diameter , it will satisfy the equation of diameter, so on putting (-3,3) in equation of diameter we get

    2(−3)−(3)+λ=0

    => −6−3−λ=0

    =>λ=9

  • Question 2
    1 / -0

    Which of the following does not defines a parabola?

  • Question 3
    1 / -0

    The equation x2+y2=0 represents

    Solution

    The above circle can be written as (x−0)2+(y−0)2=(0)2

     Here the center is (0,0) and radius is also 0 units.

    So it is a degenerate circle as  degenerate circle is a circle( a point) where radius is zero units.

  • Question 4
    1 / -0

    A circle described on the focal radii of a parabola as diameter touches the

  • Question 5
    1 / -0

    The circles x2+y2+6x+6y=0 and x2+y2−12x−12y=0

  • Question 6
    1 / -0

    If e represents the eccentricity of a parabola, then

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