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Calculus/tangent lines
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Question
Find the equations of both tangent lines to the ellipse (x^2/4)+(y^2/9)=1 that passes through the point (4,0).
I think, after all these days, that I have come up with something.
Line:
The equation is y-0 = m(x-4), so y = m(x-4)
Ellipse:
The equation given is (x/2)² + (y/3)² = 1.
Derrivative of Ellipes:
2x + 2yy' = 0, or x + yy' = 0.
Note that y' = m at the point where the line touches the ellipse.
This means that x + my = 0, so m = -x/y.
This leads to 3 equation, 3 unknowns. (x,y) is where the line intercepts the ellipse.
A) y = m(x-4) (from the line),
B) 9x² + 4y² = 36 (from the ellipse), and
C) m = -x/y.
Put (C) into (A), giving
D) y² = x(x-4).
This brings us to two equation with 2 unknowns. They are
B) 9x² + 4y² = 36 and
D) y² = x(4-x).
Putting (D) into (B) gives 9x² + 4(x(4-x)) = 36, which reduces to 9x² + 16x -4x² = 36.
WE can simplify into a quadratic equation. That is, 5x² + 16x - 36 = 0.
Using the quadratic theorem, we get x = (-16±√(256+144))/10.
Since this is an ellipse, this gives an extraneous value where x = (-16-√(256+144))/10.
The value to use is x = (-16+√(256+144))/10.
If we look at (F), we get x is about 1.5.
Putting (F) into (D) gives y is about 1.9.
From (A), we get m is about 1.7.
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