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Parabola Test - 3
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Parabola Test - 3

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  • Question 1/10
    1 / -0

    The mirror image of the parabola y2 = 4x in the tangent to the parabola at the point (1, 2) is

    Solutions

    Any point on the given parabola is (t2, 2t). 
    The equation of the tangent at (1, 2) is x - y + 1 = 0
    The image (h, k) of the point (t2, 2t) in x - y + 1 = 0 is given by

    ∴ h = t2 - t2 + 2t - 1 = 2t - 1 and
    k = 2t + t2 - 2t + 1 = t2 + 1,
    Eliminating t from h = 2t - 1 and k = t2 + 1,
    We get (h + 1)2 = 4 (k-1)
    The required equation of reflection is (x + 1)2 = 4(y - 1).

     

  • Question 2/10
    1 / -0

    AB is chord of the parabola y2 = 4ax with vertex A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is

    Solutions

     

  • Question 3/10
    1 / -0

    A focal chord of parabola y2 = 4x is inclined at an angle of  with positive x-direction, then the slope of normal drawn at the ends of chord will satisfy the equation

    Solutions

    For focal chord t1 t2 = -1,


    ⇒ t1 + t2 = 2t1 t2 = -1
    Hence the required equation m2 + 2m - 1 = 0.
    Since slope of normal drawn at A and B are -t1, -t2 respectively.

     

  • Question 4/10
    1 / -0

    If the normal at three distinct points (p2, 2p), (q2, 2q) and (r2, 2r) of the parabola y2 = 4x are concurrent then

    Solutions

    Equations of the normal are 
    px + y - 2 (p3 + 2p)=0
    qx + y - 2 (q3 + 2q) = 0
    rx + y - 2 (r3 + 2r) = 0

    ⇒ (p + q + r) (p - q)( q-r) (r - p) = 0
    ⇒ p + q + r = 0.

     

  • Question 5/10
    1 / -0

    The parabola y2 + 4x and the circle (x - 6)2 + y2 = r2 will have no common tangent if ‘r’ is equal to

    Solutions

    Any normal of parabola is y = -tx +2t + t3. 
    If it passes through (6, 0), then -6t + 2t + t3 = 0
    ⇒ t = 0, t2 = 4, A = (4,4)
    Thus for non common tangents AC > r

     

  • Question 6/10
    1 / -0

    The length of normal chord which subtend an angle of 90° at the vertex of the parabola y2 = 4x is

    Solutions

     

  • Question 7/10
    1 / -0

    From the focus of the parabola y2 = 8x, tangents are drawn to the circle (x - 6)2 + y2 = 4. Then the equation of circle through the focus and points of contact of the tangent is

    Solutions

    Focus of the parabola is (2, 0) 
    Centre of the given circle is C (6, 0) 
    Now, circle through S, P and Q will also pass through C.
    ⇒ Required circle is (x - 2) (x - 6) + y2 = 0
    ⇒ x2 + y2 - 8x + 12 = 0.

     

  • Question 8/10
    1 / -0

    Statement- 1 : The locus of the centre of circle which cuts orthogonally the parabola y2 = 4x at the point P (1, 2) is a circle.

    Statement-2 :The tangent at P to circle is a normal to the parabola at P.

    Solutions

    The tangents at P to parabola is  y.2 = 2 (x + 1) ⇒ y = x + 1.
    It must pass through the centre of the circle so the line y = x + 1 is the desired locus.

     

  • Question 9/10
    1 / -0

    The ends of a line segment are P (1, 3) and Q (1, 1). R is a point on the line segment PQ such that PR: QR = 1 : λ. If R is an interior point of the parabola y2 = 4x then

    Solutions

     

  • Question 10/10
    1 / -0

    The locus of a point such that the sum of it distances from the origin and the line x = 2 is 4 units is

    Solutions


    ⇒ y2 = 4(x + 1).

     

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