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Math Help - Coordinate Systems

  1. #1
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    Coordinate Systems

    The variable chord PQ on the parabola with equation y^2 = 4x subtends a right angle at the origin O. By taking P as (t₁^2, 2t₁) and Q as (t₂^2, 2t₂), find a relation between t₁ and t₂ and hence show that PQ passes through a fixed point on the x-axis.


    Please help me to solve this?
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  2. #2
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    Quote Originally Posted by geton View Post
    The variable chord PQ on the parabola with equation y^2 = 4x subtends a right angle at the origin O. By taking P as (t₁^2, 2t₁) and Q as (t₂^2, 2t₂), find a relation between t₁ and t₂ and hence show that PQ passes through a fixed point on the x-axis.

    Please help me to solve this?
    Let O be the origin then we have,

    (\text{slope})_{OP}\cdot (\text{slope})_{OQ} = -1

    \frac{2t_1 - 0}{t_1 ^2 - 0}\cdot \frac{2t_2 - 0}{t_2 ^2 - 0} = -1

    \color{blue}t_1 t_2 = -4

    Now the equation of the line passing through (t_1 ^2, 2t_1) and (t_2 ^2, 2t_2) is given by (t_1 + t_2)y = 2x + 2t_2t_2.

    Substitute \color{blue}t_1 t_2 = -4 to get (t_1 + t_2)y = 2x - 8.

    Clearly this line always passes through (4,0) which is on the x-axis.
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    Let O be the origin then we have,

    (\text{slope})_{OP}\cdot (\text{slope})_{OQ} = -1
    Thank you for your help. But Iíve confusion. I know that product of tangent & normal is equal to -1. So why we suppose to assume OP is tangent & OQ is normal or vice-versa?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by geton View Post
    Thank you for your help. But Iíve confusion. I know that product of tangent & normal is equal to -1. So why we suppose to assume OP is tangent & OQ is normal or vice-versa?
    No.I did not say they are tangent and normal.

    If two lines OP and OQ are perpendicular, then

    Since your question claimed "The variable chord PQ on the parabola with equation y^2 = 4x subtends a right angle at the origin O", OP and OQ are perpendicular.
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  5. #5
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    Thank you so much Isomorphism
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