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Thread: tangent to a curve

  1. #1
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    tangent to a curve

    I have problems solving this, seems a bit more complicated than other similar problems.

    For which parameter a (real number) is y=(3/2)*x-2 tangent of curve (y^2)*x+a=x^2+y^2?

    Thanks!
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  2. #2
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    Re: tangent to a curve

    Quote Originally Posted by kicma View Post
    I have problems solving this, seems a bit more complicated than other similar problems.

    For which parameter a (real number) is y=(3/2)*x-2 tangent of curve (y^2)*x+a=x^2+y^2?

    Thanks!
    Hello,

    there is probably a more accurate way to do this question - but here is what I've done:

    1. $\displaystyle (y^2)\cdot x+a=x^2+y^2~\implies~f(x,y)= x^2 +y^2-x y^2 -a = 0$

    2. By partial differentiation you'll get:

    $\displaystyle \frac{\partial f}{\partial x}=y^2-2x$.......... and............. $\displaystyle \frac{\partial f}{\partial y}=2y(x-1)$

    3. Therefore: $\displaystyle \frac{dy}{dx} = -\frac{y^2-2x}{2y(x-1)} = \frac32$

    4. Solve this equation for x:

    $\displaystyle x = \frac{y \cdot (3 - y)}{3y - 2}$

    5. From the equation of the line you'll get: $\displaystyle x = \frac23 \cdot (y+2)$

    6. Set equal those terms in x:

    $\displaystyle \frac{y \cdot (3 - y)}{3y - 2} = \frac23 \cdot (y+2)$ and solve for y.

    7. You'll get 2 values for y: y = 1 or $\displaystyle y = -\frac89$

    8. The corresponding x-values are x = 2 or $\displaystyle x = \frac{20}{27}$

    9. Plug in these values for x and y into the equation of the curve to determine a:

    $\displaystyle a = x^2+y^2-x y^2$

    $\displaystyle (x,y)=(2,1)~\implies~a= 3$ Graph in brown, tangent point in red.

    $\displaystyle (x,y)=(\frac{20}{27}\ ,\ -\frac89)~\implies~a= \frac{1648}{2187}$ Graph in blue, tangent point in violet.
    Attached Thumbnails Attached Thumbnails tangent to a curve-tanganimplizitfkt.png  
    Thanks from kicma
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