Ellipse and tangents

Question

Given the ellipse $x^2-xy+y^2=\frac{3}{4}$ I want to determine the points that their tangent is perpendicular to the $x'x$ axis and parallel to the $x'x$ axis.

Solution

The points that their tangent is perpendicular to the $x'x$ axis are:

$$A(-1, -1/2), \;\; B=(1, 1/2)$$

However how do we prove this? I set at the ellipse where $x$ the $-y$. Hence $3y^2=\frac{1}{4}$ and we get that $y=\pm 1/2$. However I don't think that this is a rigorous proof. Similarly I know that the points $A'(1/2, 1), \;\; B'(-1/2, 1)$ the tangent is parallel to the $x'x$ axis, but how do I justify this?

Answer 1

Let us consider the function $f(x,y) = x^2-xy+y^2-3/4 $. Then $$\nabla f(x,y)= \left(f_x(x,y), f_y(x,y)\right) = (2x -y, -x +2y).$$

Recall that $f_x(x,y), f_y(x,y)$ give the rate of change of $f$ in the direction of $x$ and $y$, respectively. Since we want the tangent to be perpendicular to the $x-$ axis, it is clear (?) that $$f_y(x_0,y_0) =0 \implies x_0 = 2y_0,$$ where $(x_0,y_0)$ is the point of intersection (ellipse and tangent). Since $(x_0,y_0)$ belongs to the ellipse, as well, we may solve the system: $$\left\{ \begin{array}{l} x_0 = 2y_0\\ f(x_0,y_0) = 0 \end{array} \right. $$ which yields the $2$ points $A(-1,-1/2)$ and $B(1,1/2)$.

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The concept is the same when the tangent is parallel to the $x-$ axis (which means that the tangent is perpendicular to the $y-$ axis).

Answer 2

By rotating the equation:

$$\frac{x^{2}}{\frac{3}{2}}+\frac{y^{2}}{\frac{1}{2}}=1$$

Now to find the new A and B coordinates:

$x'=x\cos \left( \theta \right)+y\sin \left( \theta \right)$

$y'=y\cos \left( \theta \right)-x\sin \left( \theta \right)$

where theta is how much you rotated clockwise (here π/4 or 45º).

$$A_{r}\left( \frac{-3}{2\sqrt{2}},\frac{1}{2\sqrt{2}} \right)\; and\; B_{r}\left( \frac{3}{2\sqrt{2}},\frac{-1}{2\sqrt{2}} \right)$$

Now when you find the tangents at the new A and B points you will get $$y=x+\sqrt{2}\; and\; y=x-\sqrt{2}$$

The angle of the slope is 45º, so rotate back 45º to the original function and you'll get 90º.

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I do this calculus way, which could be wrong but here:

Through implicit differentiation:

$$\frac{dy}{dx}=\frac{2x-y}{x-2y}$$

If we put the points $A\left( -1,-\frac{1}{2} \right)$ and $B\left( 1,\frac{1}{2} \right)$ we get, respectively: $$\frac{dy}{dx}=\frac{-2+\frac{1}{2}}{0}\; and\; \frac{dy}{dx}=\frac{2-\frac{1}{2}}{0}$$

So we can see that the slope is infinite (parallel to the y-axis), also $\frac{dx}{dy}=0$ at those points, not $\frac{0}{0}$ or something undefined; so this proves that the tangents at $A$ and $B$ are perpendicular to the x-axis.

Answer 3

If $x^2-xy+y^2=\frac{3}{4} $, then $0 =2xdx-(xdy+ydx)+2ydy =(2x-y)dx+(2y-x)dy $, so $\frac{dy}{dx} =-\frac{2x-y}{2y-x} $ and $\frac{dx}{dy} =-\frac{2y-x}{2x-y} $.

If $dy/dx = 0$, $y = 2x$ so $\frac34 =x^2-2x^2+4x^2 =3x^2 $ or $x=\pm\frac12$ and $y = \pm 1 $, with the same sign for $x$ and $y$.

If $dx/dy = 0$, $y = x/2$ so $\frac34 =x^2-x^2/2+x^2/4 =3x^2/4 $ or $x=\pm 1$ and $y = \pm 1/2 $, with the same sign for $x$ and $y$.