Question

The equation\({x^2} - xy + {y^2} = 3\) represents a “rotated ellipse,” that is, an ellipse whose axes are not parallel to the coordinate axes. Find the points at which this ellipse crosses the \(x - \)axis and show that the tangent lines at these points are parallel.

Short answer

The slopes of the tangent at both points are \(2\)so they are parallel.

Step-by-step solution

Given Information.

The given expression is\({x^2} - xy + {y^2} = 3.\)

Meaning of Ellipse.

An ellipse is the locus of all those points in a plane such that the sum of their distances from two fixed points in the plane, is constant.

Find the points at which ellipse crosses the \(x - \)axis.

It intersects the \(x - \)axis by plugging in \(y = 0\)

\(\begin{array}{l}{x^2} - xy + {y^2} = 3\\{x^2} - x(0) + {0^2} = 3\\{x^2} = 3\\x = \pm \sqrt 3 \end{array}\)

It crosses the\(x - \)axis at \(( - \sqrt 3 ,0),(\sqrt 3 ,0)\)

Implicitly differentiate, solve for\({y^\prime }\)

\(\begin{array}{*{20}{r}}{{x^2} - xy + {y^2} = 3}\\{2x - \underbrace {\left( {{x^\prime }y + x{y^\prime }} \right)}_{{\rm{product rule }}} + 2y{y^\prime } = 0}\\{2x - \left( {y + x{y^\prime }} \right) + 2y{y^\prime } = 0}\\{2x - y - x{y^\prime } + 2y{y^\prime } = 0}\end{array}\)

Further solve

\(\begin{array}{l}2y{y^\prime } - x{y^\prime } = y - 2x\\{y^\prime }(2y - x) = y - 2x\\{y^\prime } = \frac{{y - 2x}}{{2y - x}}\end{array}\)

Show that the tangent lines at these points are parallel.

Plug the\(2\)points of the intersection into \({y^\prime }\)

Put\(x = - \sqrt 3 \)into\({y^\prime }\)

\(\begin{array}{l}{y^\prime } = \frac{{0 - 2( - \sqrt 3 )}}{{2(0) - ( - \sqrt 3 )}}\\\frac{{2\sqrt 3 }}{{\sqrt 3 }} = 2\end{array}\)

Put\(x = \sqrt 3 \)into\({y^\prime }\)

\(\begin{array}{l}{y^\prime } = \frac{{0 - 2(\sqrt 3 )}}{{2(0) - (\sqrt 3 )}}\\\frac{{ - 2\sqrt 3 }}{{ - \sqrt 3 }} = 2\end{array}\)

The slope of the tangent at both points are \(2,\)so they are parallel.