What is the shortest distance from the ellipse $x^2+\frac{y^2}{4} =1$ and the point (2,2).

Question

What is the shortest distance from the ellipse $x^2+\frac{y^2}{4} =1$ and the point (2,2).

I know that this is the projection of (2,2) onto the ellipse. How do I go about finding the shortest distance between this point and the curve?

Answer 1

You can think about this as a constrained optimization problem. You want to minimize the function $f(x,y) = (x-2)^2 + (y-2)^2$ [the square of the distance from $(x,y)$ to $(2,2)$], subject to the constraint that $x^2+y^2/4=1$.

Answer 2

Define point $P(2,2)$. The tangents to the curve $$x^2+\frac14y^2=1$$ will have a slope of $m=-\dfrac{4x}{y}$ or $-\dfrac{2x}{\sqrt{1-x^2}}$. At some $x$ value, the point of tangency will be $Q(x,2\sqrt{1-x^2})$. We want $m$ and $PQ$ to be perpendicular, and thus their values will be of negative reciprocal.

As such, we have $$\frac{\sqrt{1-x^2}}{2x}=\frac{2\sqrt{1-x^2}-2}{x-2}\implies x\approx0.69661$$

(use a numerical method).

Now that you have your point $Q$, you can find the length of $PQ$.

Answer 3

Knowing that the gradients of level curves are perpendicular to them then seek for $(x,y)$ such that $$[2x,\frac{1}{2}y]=\lambda[2(x-2),2(y-2)].$$ This models how the normal to the ellipse is aligned to the normal of a circle centered at $(2,2)$.

The next plot may help you to think towards the solution

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