Find the points at which the ellipse x2−xy+y2=3 crosses the x-axis and show that the tangent lines at these points are parallel.
The ellipse crosses the x-axis at y=0 so...
x2−xy+y2=3x2−x(0)+(0)2=3x2=3x=±√3
Taking the derivative of the curve implicitly we have,
x2−xy+y2=3
ddx(x2)−ddx(xy)+ddx(y2)=ddx(3)2x−[ddx(x)⋅y+x⋅ddx(y)]+ddy(y2)dydx=02x−[y+xdydx]+2ydydx=02x−y−xdydx+2ydydx=0dydx=2x−yx−2y
@ x=√3
dydx=2(√3)−0√3−2(0)dydx=2\cancel√3\cancel√3dydx=2
@ x=−√3
dydx=2(−√3)−0−√3−2(0)dydx=2(\cancel−√3)\cancel−√3dydx=2
The slopes are equal. Therefore, the tangent lines are parallel.
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