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:
$\displaystyle x = \frac{y \cdot (3 - y)}{3·y - 2}$
5. From the equation of the line you'll get: $\displaystyle x = \frac23 \cdot (y+2)$
6. Set equal those terms in x:
$\displaystyle \frac{y \cdot (3 - y)}{3·y - 2} = \frac23 \cdot (y+2)$ 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.