posted by dt
find the lines that are (a) tangent and (b) normal to the curve at the given point
x^2 + xy - y^2 = 1
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
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.
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)
4) x - 4y = -31
2x - 4y = -34