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Conic Sections Test - 5

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Conic Sections Test - 5
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Weekly Quiz Competition
  • Question 1
    1 / -0

    A and B are two distinct points, Locus of a point P satisfying |PA| + | PB | = 2k, a constant is

    Solution

    Since the value of k is not specified, we cannot get a specific equation.

  • Question 2
    1 / -0

    locus of the point of intersection of the lines x = sec θ + tan θ and y = sec θ – tan θ is

    Solution

    After solving the equations we will get x+y = 2sec θ which represents a linear equation. 

  • Question 3
    1 / -0

    The vertex of the parabola y2 = 4 a(x − a) is

    Solution

    y2 = 4 a(x − a)

    Let Y=y and X = x-a

    for standard parbola Y2=4aX the vertex is (0,0)

    so put X=0 and Y=0.

    i.e. put x-a = 0 and y=0

    so vertex is (a,0)

  • Question 4
    1 / -0

    The two parabolas x2 = 4y and y2 = 4x meet in two distinct points. One of these is the origin and the other is

    Solution

    substituting for x2 in equation (2)

    y2 = 4y. This implies y = 4

    similarly substituting for y2 in equation(1) 

    x2 = 4x. This implies x = 4

    Hence the other point point of intersection is (4,4)

  • Question 5
    1 / -0

    The equation of the directrix of the parabola x2 = − 4ay is

    Solution

    From the equation we infer that the parabola is open downward.

    Hence the directrix passes through the point (0,a) and will be parallel to the X-axis. 

    Hence the equation of the directrix is y = a or y - a = 0 

  • Question 6
    1 / -0

    The eccentricity ‘e’ of a parabola is

    Solution

    Eccentricity of a parabola is 1

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