Expert: Paul Klarreich - 12/14/2008

Question
Consider the curve given by xy^2 - (x^3)y = 6
a. Show that dy/dx = (3(x^2)y - y^2) / (2xy - x^3)
b. Find all points on the curve whose x-coordinate is 1,
and write an equation for the tangent line at each of these
points
c. find the x - coordinate of each point on the curve where
the tangent line is vertical

I have tried everything that we learned, taking the
derivative implicitly and so forth, but I havent been able
to get past a. and even that was wrong.

Answer
Questioner:   Chris
Category:  Advanced Math
Private:  No
 
Subject:  Calculus
Question:  Consider the curve given by xy^2 - (x^3)y = 6
a. Show that dy/dx = (3(x^2)y - y^2) / (2xy - x^3)
b. Find all points on the curve whose x-coordinate is 1,
and write an equation for the tangent line at each of these
points
c. find the x - coordinate of each point on the curve where
the tangent line is vertical

I have tried everything that we learned, taking the
derivative implicitly and so forth, but I havent been able
to get past a. and even that was wrong.
...........................
Hi, Chris,

Did you actually get (a)?  It looks right to me.
...............
b. Set  x = 1 and solve for y.

(1)y^2 - (1^3)y = 6

y^2 - y = 6

Solve this quadratic equation. (Looks easy enough.)

NOW you have two points -- two sets of coordinates (x,y).

Just plug those into your dy/dx to get m.  Now use the point-slope form:

y - y0 = m(x - x0).

You have two sets of x0,y0,m, so you get two equations.

........................

c. If the t.l. is vertical, then m is undefined.  So look for a value of x that makes the denominator of your dy/dx equal to zero.