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# ?

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Answer 1 · 2017-12-15T03:21:31

#p=1, q=-5/2#

I have made a video that outlines all of my steps :)

I hope it helps :)

Harold

Answer 2 · 2017-12-15T10:39:13

#p = 1, q = -5/2#

Considering

#f(x,y) = y -px^3-qx=0# and
#g(x,y) = y-19/2x+16=0#

we have

#nabla f = (-3px^2-q,1)# and
#nabla g = (-19/2,1)#

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

At point #p_0 = (x_0,y_0)# we have #nabla f = nabla g#

#{(-3px_0^2-q = -19/2),(1=1),(y_0-px_0^3-q x_0=0):}#

solving for #p,q#

#{(p = (19 x_0-2y_0)/(4x_0^2)),(q = -19/4+(3y_0)/(2x_0)):}#

which at #p_0# are #p = 1, q = -5/2#