Question

The gradient of the curve y-xy+2px+3qy=0 at the point (3,2) is -2/3. the value of p and q are

Answer 1

The gradient of the curve $y - xy + 2px + 3qy = 0$ at the point $(3,2)$ is $-2/3$. The value of $p$ and $q$ are?

**Solution:**

We have

$$y - x y + 2 p x + 3 q y = 0.$$

Thus

$$\frac {d}{d x} (y - x y + 2 p x + 3 q y) = 0,$$ $$\frac {d y}{d x} - x \frac {d y}{d x} - y + 2 p + 3 q \frac {d y}{d x} = 0.$$

We have two conditions

$$y | _ {x = 3} = 2$$

and

$$\left. \frac {d y}{d x} \right| _ {\substack {x = 3 \\ y = 2}} = - \frac {2}{3}.$$

So we get system of two equations

$$\left\{ \begin{array}{c} 2 - 3 \cdot 2 + 2 p \cdot 3 + 3 q \cdot 2 = 0, \\ - \frac {2}{3} - 3 \cdot \left(- \frac {2}{3}\right) - 2 + 2 p + 3 q \cdot \left(- \frac {2}{3}\right) = 0, \end{array} \right.$$ $$\left\{ \begin{array}{l} 3 p + 3 q = 2, \\ p - q = \frac {1}{3}, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} p + q = \frac {2}{3}, \\ p - q = \frac {1}{3}, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 2 q = \frac {1}{3}, \\ 2 p = 1, \end{array} \right. \Rightarrow \left\{ \begin{array}{l} q = \frac {1}{6}, \\ p = \frac {1}{2}. \end{array} \right.$$

**Answer:**

$$q = \frac {1}{6}$$ $$p = \frac {1}{2}$$