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Math Help - Finding the eq of a normal to the parabola

  1. #1
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    Finding the eq of a normal to the parabola

    Find the eq of the normal at a pt on the parabola y^2 = 2x whose ordinate is 4

    Solution

    Eq of tangent to parabola at (x_1,y_1) is \Rightarrow yy_1 = 2a(x+x_1)

    Eq of parabola is y^2 = 2x \therefore a = \frac{1}{2}

    \therefore the eq of a tangent to the parabola at (x_1,4) is \Rightarrow 4y = (x+x_1) \Rightarrow y = \frac{1}{4}(x+x_1)

    since the normal i perpendicular to the tangent then its slope would be -4

    \therefore eq of the normal to the parabola at a pt (x_1, 4) is

    y = -4 (x+x_1)

    Am i correct !!!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by flintstone View Post
    Find the eq of the normal at a pt on the parabola y^2 = 2x whose ordinate is 4

    Solution

    Eq of tangent to parabola at (x_1,y_1) is \Rightarrow yy_1 = 2a(x+x_1)

    Eq of parabola is y^2 = 2x \therefore a = \frac{1}{2}

    \therefore the eq of a tangent to the parabola at (x_1,4) is \Rightarrow 4y = (x+x_1) \Rightarrow y = \frac{1}{4}(x+x_1)

    since the normal i perpendicular to the tangent then its slope would be -4

    \therefore eq of the normal to the parabola at a pt (x_1, 4) is

    y = -4 (x+x_1)

    Am i correct !!!!!
    No. and the fact that you have x_1 in your answer makes this obvious. the only variables should be x and y.

    y^2 = 2x. When y = 4, x = 8, so the point the normal line passes through is (8,4). Now we need to find the slope.

    Differentiating implicitly we see that y' = \frac 1y. Which means the slope when y = 4 is 1/4. which implies the slope of the normal line is -4.

    I think you can take it from here.

    (did you make an error? should this be in precalc or calc?)
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  3. #3
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    thanks for the reply
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