SOLUTION: what is the equation of the tangent line and normal line of to the ellipse x^2+2y^2=9 perpendicular to the line 4x-y=6

Question 990045: what is the equation of the tangent line and normal line of to the ellipse x^2+2y^2=9 perpendicular to the line 4x-y=6
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
what is the equation of the tangent line and normal line of to the ellipse perpendicular to the line 4x-y=6
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The slope of 4x-y = 6 is +4
The slope of lines perpendicular to it is -1/4.
Find the points on the ellipse with those slopes.
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x^2+2y^2=9
2xdx + 4ydy = 0
dy/dx = -x/2y
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For the 2 points with a slope of +4:
x^2 + 2y^2 = 9
-x/2y = 4
x = -8y
(-8y)^2 + 2y^2 = 9
y^2 = 9/66

x = -8y = -4sqrt(66)/11
--> (-4sqrt(66)/11,+sqrt(66)/22)
and (4sqrt(66)/11,-sqrt(66)/22)
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With the slope and the points known, can you find the 2 equations?
Then do the same for the perpendicular lines.
email via the TY note for help or to check your work.

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