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Tangent to ellipse

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2 I have done question a, ,my answers are a=2, b=-2 c=141 please correct if im wrong

Please help me with with question b as i have no idea how to do it

Oct 17, 2018

#2
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Type error my answer for a was correct XDD

Thank you so much once again!!!!!

Oct 17, 2018

#1
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(a)

We can transform the original equation to this :

4x^2  + 25y^2   = 100

The center of this ellipse is  at  (0,0)

So..the translated equation is

4x^2 - 16x  + 25y^2 + 150y  =   -c     complete the square on  x and y

4(x^2 -  4x + 4)  + 25 (y^2 + 6y + 9)  =  -c + 16 + 225

4(x - 2)^2  + 25(y +3 )^2  = 241  - c       divide through by 100

(x - 2)^2 / 25 + (y +3 ) / 4  =    [241 - c ] / 100

We want the right side to =  1...so....c  = 141

The center of this ellipse is at  (2 , -3)   so....this is the transaltion  vector

So  a  = 2   b  = -3    and c  = 141

(b)  The center of the translated ellipse is (2, -3)

The minor axis  is parallel to the y axis

The length of this axis  = 4

And the tangent lines tangent to the ellipse and parallel to the x axis  are 2 units above and below the center.....so....their equations are  y = -3 + 2   ⇒  y  = -1  and

y = -3 - 2  ⇒ y = -5

Here's the graph : https://www.desmos.com/calculator/k2p2x4xh7k   Oct 17, 2018
#2
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