x^2 + 2xy - 3y^2 = 0 The line that is normal to this curve at (1,1) intersects the curve at what other point? I got the slope to be 1. Can someone tell me how to do this? Please?
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Originally Posted by Morgan82 x^2 + 2xy - 3y^2 = 0 The line that is normal to this curve at (1,1) intersects the curve at what other point? I got the slope to be 1. Can someone tell me how to do this? Please? normal slope is -1 $\displaystyle y-1 = -1(x-1)$ $\displaystyle y = -x + 2$ substitute (-x+2) in for y in the original equation ... solve for x, then determine y
If I plug it back in, am I supposed to get (-16+- sqrt(192)) / 8? Im still confused.
Originally Posted by Morgan82 If I plug it back in, am I supposed to get (-16+- sqrt(192)) / 8? Im still confused. no ... you should get x = two very nice integer solutions.
Which equation do I plug it into?
Originally Posted by skeeter substitute (-x+2) in for y in the original equation ... ..
oh I got it... a dumb mistake in the solving. Thankyou!
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