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

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

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Answer 1 · 2017-12-15T03:21:31

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

Explanation:

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Harold

Answer 2 · 2017-12-15T10:39:13

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

Explanation:

Considering

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

we have

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

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

At point $p_{0} = (x_{0}, y_{0})$ $p_0 = (x_0,y_0)$ we have $\nabla f = \nabla g$ $nabla f = nabla g$

$⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} - 3 p x_{0}^{2} - q = - \frac{19}{2} 1 = 1 y_{0} - p x_{0}^{3} - q x_{0} = 0 \end{matrix}$ ${(-3px_0^2-q = -19/2),(1=1),(y_0-px_0^3-q x_0=0):}$

solving for $p, q$ $p,q$

$⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} p = \frac{19 x_{0} - 2 y_{0}}{4 x_{0}^{2}} q = - \frac{19}{4} + \frac{3 y_{0}}{2 x_{0}} \end{matrix}$ ${(p = (19 x_0-2y_0)/(4x_0^2)),(q = -19/4+(3y_0)/(2x_0)):}$

which at $p_{0}$ $p_0$ are $p = 1, q = - \frac{5}{2}$ $p = 1, q = -5/2$

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

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If 19x−2y−32=019x - 2y - 32 = 0 is a tangent to the curve y=px3+qxy = px^3 + qx at (2,3)(2,3). Then what are the values of pp and qq ?

2 Answers

Explanation:

Explanation:

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