The two equations I have are 9x^2 + 25y^2=169 (nine x squared plus 25 y squared = 169) and x^2+y^2+28x-23y=152.
I have to prove they touch at (4,-1) and no where else. The first part (that they meet at that point) is easy enough, but what about the second bit?
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x^2/(169/9) + y^2/(169/25) = 1
this is an ellipse with center at (0,0) ... the possible x-values are from -13/3 to 13/3 ... the possible y-values are from -13/5 to 13/5
x^2 + 28x + 196 + y^2 - 23y + 529/4 = 152 + 196 + 529/4 = 340.25
this is a circle with center at (-14 , 11.5) with radius...18.44.
it turns out that the ellipse is totally inside the circle...
there will be one intersection if their tangent values at the point are the same...
for ellipse... 9x + 25y dy/dx = 0 ... dy/dx = -36/25
for circle .. 2x + 2y dy/dx + 28 - 23 dy/dx = 0
dy/dx = -(2x+28)/(23 - 2y) = -36/25
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the both equations represents two circles
the first eqn gives circle with origin as center and the second one is having center (14, -11.5) if i am not wrong.
now check the the conditons such the circles touch with each other...
as it is lng process to solve here..i gave u clue how to solve it out..just go along
Maybe graphically, by drawing the two shapes?
They're an ellipsoid and a circle, it can't be too hard.
Ask your professor or tuition teacher.
i hate equations..