Question

Consider the curve defined by 2y^2-x^2y=3.

a. Show that dy/dx=2xy/3y-x^2

b. Write an equation for the line tangent to the curve at the point (1, -1).

c. Show that there is a point P with x=coordinate 0 at which the line tangent to the curve P is horizontal. Then find the y coordinate of P.

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Answer 1

(a)

$$\left.\frac{d}{dx}\right|2y^2-x^2\cdot y =3\to 2\cdot 2y\cdot y'-2x\cdot y-x^2\cdot y'=0\\[0.3cm] \left.\left(4y-x^2\right)\cdot y'=2xy\right|\div\left(4y-x^2\right)\\[0.3cm] \boxed{y'=\frac{2xy}{4y-x^2}}$$

Q.E.D.

(b)

Note : according to the condition of the problem, we need to write the equation of the tangent to the point $P(1,-1)$ - this means that $x_0=1\,\,\,\text{and}\,\,\,y(1)=-1$ .

From the theory we know that the equation of the tangent line at a point $x=x_0$ has the form

$$y_{tang}(x)=y(x_0)+y'(x_0)\cdot(x-x_0)$$

In our case,

$$x_0=1\to y(1)=-1\to y'(1)=\frac{2\cdot1\cdot y(1)}{4\cdot y(1)-1^2}=\\[0.3cm] =\frac{2\cdot(-1)}{4\cdot(-1)-1}=\frac{-2}{-4-1}=\frac{2}{5}\longrightarrow\\[0.3cm] y_{tang}(x)=y(1)+y'(1)\cdot(x-1)=-1+\frac{2}{5}\cdot(x-1)=\\[0.3cm] =-\frac{5}{5}+\frac{2x}{5}-\frac{2}{5}=\frac{2x}{5}-\frac{7}{5}\\[0.3cm] \boxed{y_{tang}(x)=\frac{2x}{5}-\frac{7}{5}\,\,\,,\,\,\,\text{at}\,\,\,x=1}$$

(c)

Note : We have to show that the tangent at $x=0$ is horizontal - this means that the derivative is 0. To do this, first, we find the value of the function at $x=0$ . Let's use the equation

$$\left\{\begin{array}{l} x=0\\[0.3cm] 2y^2-x^2\cdot y=3 \end{array}\right.\to2y^2-0^2\cdot y=3\to 2y^2=3\\[0.3 cm] \boxed{y(0)=\pm\sqrt{\frac{3}{2}}}$$

Note : For our purpose, it is enough to understand that $y(0)=y_0\neq0$ . Then,

$$\text{from part (a)}\,\,\, :\,\,\,y'(x)=\frac{2xy}{4y-x^2}\to y'(0)=\frac{2\cdot0\cdot\overbrace{y(0)}^{y_0}}{4\cdot\underbrace{y(0)}_{y_0}-0^2}=0\\[0.3cm] \boxed{y'(0)=0\longrightarrow\text{the tangent is horizontal}}$$

Q.E.D.

ANSWER

$$y_{tang}(x)=\frac{2x}{5}-\frac{7}{5}\,\,\,,\,\,\,\text{at}\,\,\,x=1$$