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Determine the equations for all lines tangent to the ellipse

$$ f(x,y)=x^2+xy+y^2=1 $$

and passing through the point (0,2)

I'm a bit stuck on this question. I was thinking if I have an unknown point on the curve called (a.b). Then I can plug it in the curve. $$f(a.b)=a^2+ab+b^2=1$$ And if I also find the derivative of it, which is: $$(2a+b,a+2b) $$ But from here I really don't know how to proceed. Maybe if I can somehow set it up as simultaneous equations and find the unknown point, from there I can easily find the tangent line.

karl johan
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4 Answers4

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Note that a parameterization for your ellipse is

$$ \begin{align} x(t)&= -\cos(t)+\frac{1}{\sqrt{3}}\sin(t)\\\\y(t)&= \cos(t)+\frac{1}{\sqrt{3}}\sin(t) \text{ .} \end{align}$$

The tangent (velocity) vector at any point would then be

$$ \vec{v} = \begin{bmatrix} x'(t)\\y'(t) \end{bmatrix} = \begin{bmatrix} \sin(t) - \frac{1}{\sqrt{3}}\cos(t)\\-\sin(t) - \frac{1}{\sqrt{3}}\cos(t) \end{bmatrix} \text{ .}$$

For a particular point along the parameterization $\vec{p} = (x(\tau),y(\tau))$ on the ellipse, the tangent line would be given by

$$ \vec{x} = \vec{p} + \vec{v}(\tau)t $$

where $\vec{x} = (x_\ell(t),y_\ell(t))$ is a parameterization for the tangent line.

Let me know if you have any follow-up questions.

vb628
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Here is one approach:

$x^2+xy+y^2=1$

Taking derivative wrt. $x$, $2x + y + x y' + 2y y' = 0 \implies y' = -\frac{2x+y}{x+2y}$

This is the slope of the tangent line at a point $T (x, y)$ on ellipse. As the tangent line goes through point $P (0, 2)$, we can equate slope of $PT$ to $y'$.

$-\frac{2x+y}{x+2y} = \frac{y-2}{x} \implies x^2+xy+y^2 = x + 2y$.

As point $T (x, y)$ is on ellipse, $x^2 + xy + y^2 = 1$

So, $x = 1 - 2y$ and plugging it into equation of ellipse,

$(1-2y)^2 + y(1-2y) + y^2 = 1 \implies y(y-1) = 0$

That gives us two points on the ellipse $(1, 0)$ and $(-1, 1)$. Tangent lines at these points pass through point $P(0, 2)$.

You can now find the equation of tangent lines.

Math Lover
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A general line through $(0,2)$ would be parametrised as $$\mathbf{r}(t)=(0,2)+t(\alpha,\beta).$$ Substitute that into the equation for the ellipse and then you will have an expression which you can treat as a quadratic in t. The line will be tangential to the ellipse if and only if there is just one solution for t i.e. discriminant $b^2-4ac=0$. This should narrow down the possible choices for the direction $(\alpha,\beta)$ to only two options. Hope this helps

jcneek
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Extremely easy using the trick of homogenization :

$$x^2 +y^2 + xy -1=0 \tag{1}$$

Homogenize.

$$ x^2 +y^2 + xy -z^2=0$$

Now, add in some mindless derivatives:

$$ X(2x +y) + Y(2y +x) +(-2z)=0$$

What does above represent? It represents of the chord of contact from the tangents drawn to an external point $P=(X,Y)$

Evaluate at z=1,

$$ X(2x+y) + Y(2y+x) - 2= 0 \tag{2}$$

Now, plug in $(0,2)=(X,Y)$ $$ 2(2y+x) -2 = 0$$

$$ 2y+x=1 \tag{3}$$

Now, simply find the intersection of the above equation with $(2)$, this will lead you to getting some $(x,y)$ pair which is a point on the curve. Each of this point corresponds to a tangent line you can draw from curve to external point, to convert the point to tangent line, plug the $(x,y)$ pair into $(2)$.

I find these pairs to be $(x,y) = \{ (-1,1), (1,0)\}$

Reference for method

tryst with freedom
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