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Conic Sections Test - 7

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Conic Sections Test - 7
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  • Question 1
    1 / -0

    The line y = mx + 1 is a tangent to the parabola y2 = 4xy2 = 4x if m =

    Solution

    The contion for a line to be a tangent to the parabola is y = mx +a/m

    From the given equation of the parabola we infer that a = 1 and from the equation of the line c = 1

    Therefore m = a/c = 1

  • Question 2
    1 / -0

    The equation x2 − 7xy + 12y2 = 0 represents a

    Solution

    On factorizing the given equation we get x-4y=0 and x-3y=0, which represents a pair of intersecting straight lines. Since the product of their slopes is not equal to -1, they are non perpendicular.

  • Question 3
    1 / -0

    The equation y2 − x2 + 2x − 1 = 0 represents

    Solution

    On factorizing the given equation we get y+x-1=0 and y-x+1=0, which clearly represent a pair of straight lines which are intersecting.

  • Question 4
    1 / -0

    S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is

    Solution

    s = (ae,0) and T = (-ae,0) and B = (0,b)

    Since it is an equilateral triangle, ST2 = TB2

    This implies 4a2e2 = a2e2 + b2

    3a2e2 = b2 

    3a2e2 = a2(1 - e2)

    3e2  = 1 - e2

    Therefore e = 1/2

  • Question 5
    1 / -0

    The slope of the tangent at the point (a, a) of the circle x2 + y2 = a2 is

    Solution

    the equation of the tangent is x+y=a. Hence its slope is -1

  • Question 6
    1 / -0

    The equation of the normal to the parabola y2 = 8 having slope 1 is

    Solution

    slope form of normal is y=mx-2am-am3

    for the given parabola a = 2 and m = 1

    therefore y = x -4 -2 

    i.e; x - y - 6 = 0

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