# Math Help - tangent to a curve

1. ## 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!

2. ## Re: tangent to a curve

Originally Posted by kicma
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. $(y^2)\cdot x+a=x^2+y^2~\implies~f(x,y)= x^2 +y^2-x y^2 -a = 0$

2. By partial differentiation you'll get:

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

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

4. Solve this equation for x:

$x = \frac{y \cdot (3 - y)}{3·y - 2}$

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

6. Set equal those terms in x:

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

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

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

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

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

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

$(x,y)=(\frac{20}{27}\ ,\ -\frac89)~\implies~a= \frac{1648}{2187}$ Graph in blue, tangent point in violet.