Asked by Jane
Find:
1)x^2-y^2=15, normal at (-8,7)
1)x^2-y^2=15, normal at (-8,7)
Answers
Answered by
bobpursley
first, find the slope, then the negative reciprocal of that slope.
2x dx-2ydy=0
Put in x,y and find dy/dx.
Now, the normal will be y=mx+b, m is the negative reciprocal of the dy/dx above, solve for b, and you have it.
2x dx-2ydy=0
Put in x,y and find dy/dx.
Now, the normal will be y=mx+b, m is the negative reciprocal of the dy/dx above, solve for b, and you have it.
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