# If 19x - 2y - 32 = 0 is a tangent to the curve y = px^3 + qx at (2,3). Then what are the values of p and q ?

Dec 15, 2017

$p = 1 , q = - \frac{5}{2}$

#### Explanation:

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I hope it helps :)

Harold

Dec 15, 2017

$p = 1 , q = - \frac{5}{2}$

#### Explanation:

Considering

$f \left(x , y\right) = y - p {x}^{3} - q x = 0$ and
$g \left(x , y\right) = y - \frac{19}{2} x + 16 = 0$

we have

$\nabla f = \left(- 3 p {x}^{2} - q , 1\right)$ and
$\nabla g = \left(- \frac{19}{2} , 1\right)$

the surfaces normal vectors for $f$ and $g$

At point ${p}_{0} = \left({x}_{0} , {y}_{0}\right)$ we have $\nabla f = \nabla g$

$\left\{\begin{matrix}- 3 p {x}_{0}^{2} - q = - \frac{19}{2} \\ 1 = 1 \\ {y}_{0} - p {x}_{0}^{3} - q {x}_{0} = 0\end{matrix}\right.$

solving for $p , q$

$\left\{\begin{matrix}p = \frac{19 {x}_{0} - 2 {y}_{0}}{4 {x}_{0}^{2}} \\ q = - \frac{19}{4} + \frac{3 {y}_{0}}{2 {x}_{0}}\end{matrix}\right.$

which at ${p}_{0}$ are $p = 1 , q = - \frac{5}{2}$