Posted by dt on Sunday, October 19, 2008 at 7:23pm.
find the lines that are (a) tangent and (b) normal to the curve at teh given point
x^2 + xy - y^2 = 1
- 12th grade calculus - bobpursley, Sunday, October 19, 2008 at 7:58pm
Take the differential...
2xdx+ y dx + x dy -2ydy=0
solve for dy/dx
Then use that as m in
y=mx + b Puttingin the x,y point, solve for b.
then, for the normal, take the negative reciprocal of m, and again, solve for the line
y=-1/m x + b
- 12th grade calculus - Reiny, Sunday, October 19, 2008 at 7:58pm
You did not state the given point.
Using implicit derivative I found it to be
y' = (2x+y)/(2y-x)
sub in the given point, that gives you the slope of the tangent.
Now that you have the slope (m) and a given point, use the grade 9 method of finding the equation of the tangent.
Take the negative reciprocal of your slope, and the given point to find the equation of the normal.
- 12th grade calculus - Damon, Sunday, October 19, 2008 at 8:00pm
You forgot to say what point.
However in general we find the derivative, dy/dx and call that the slope, m, of the tangent at the point. Then plug the x and y of the point in to get b in y = m x + b
to get the derivative
2 x dx/dx +x dy/dx + y dx/dx - 2 y dy/dx = 0
2x + y = (2y-x) dy/dx
dy/dx = (2x+y)/(2y-x)
- 12th grade calculus - -, Sunday, October 19, 2008 at 8:13pm
- 12th grade calculus - ralonda, Monday, April 12, 2010 at 5:16pm
4) x - 4y = -31
2x - 4y = -34
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