find the lines that are (a) tangent and (b) normal to the curve at the given point

x^2 + xy - y^2 = 1

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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

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2. 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.

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posted by Reiny
3. 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
or
2x + y = (2y-x) dy/dx
dy/dx = (2x+y)/(2y-x)

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posted by Damon
4. thank alot

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posted by -
5. 4) x - 4y = -31
2x - 4y = -34

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posted by ralonda

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