SOLUTION: Find the equation of the tangent and normal to the conic 4x^2 + 9y^2 = 40 at point (1,-2).

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Question 380177: Find the equation of the tangent and normal to the conic 4x^2 + 9y^2 = 40 at point (1,-2).
Found 2 solutions by richard1234, robertb:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Given the ellipse 4x%5E2+%2B+9y%5E2+=+40, if we differentiate with respect to x, we have
8x+%2B+18y%28dy%2Fdx%29+=+0
dy%2Fdx+=+-8+x%2F18+y+=+-4+x%2F9+y
Therefore the slope at (1, -2) is %28-4%2A1%29%2F%289%2A-2%29+=+2%2F9
Given slope 2/9 and a point (1, -2) we can easily find the y-intercept:
-2+=+%282%2F9%29%281%29+%2B+b
b+=+-20%2F9
Thus the equation of the line tangent to the ellipse at (1, -2) is y+=+%282%2F9%29x+-+20%2F9

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The equation of the normal line: From the slope of the tangent line above at the specified point, 2/9, the slope of the normal line should be -9/2. The equation of the normal line should then be y+-+-2+=+%28-9%2F2%29%28x-1%29, or y+=+-9x%2F2+%2B+5%2F2 after simplification.