Shortest distance to a straight line

Question

Find the coordinates of the point on the ellipse $4x^{2}+y^{2}=4$, which is closest to the straight line $x+y=10$. I could solve it by using Lagrange Multiplier. Is there any way to solve it graphically?

Thank you.

Answer 1

The Lagrange multiplier method has a nice graphical interpretation.

We look for points where the normals are parallel.

The normal to the line is $(1,1)^T$.

The normal to the ellipse at $(x,y)$ is given by $(8x,2x)^T$.

Hence we look for points (x,y) lying on the ellipse satisfying $(8x,2x)^T= \lambda (1,1)^T$.

This gives ${5 \over 16} \lambda^2 = 4$, or $\lambda = \pm \sqrt{64 \over 5}$, from which we can get the $(x,y)$ values.

Alternative:

Consider the family of parallel lines $L_t$ given by $x+y=t$ and find the $t$ for which the intersection of the ellipse and the line $L_t$ results in exactly one point.

Substituting $y=t=x$ into the equation for the ellipse gives $4x^2+(t-x)^2 =4$ and solving for $x$ gives $x = { t \pm 2 \sqrt{5-t^2} \over 5} $. Hence we must have $t = \pm \sqrt{5}$. Substituting gives $x$ and then $y = t-x$.

Answer 2

Is there any way to solve it graphically?

I measured it in GeoGebra. It is about $(0.39,1.84)^t$.

(Large version)

Geometrical Solution:

The line $a$ was $$ (x, y)^T \cdot (1/\sqrt{2}, 1/\sqrt{2})^T = 10/\sqrt{2} $$ A normal line $b$ to it, with intersection $(x,y)^T$ on $a$ is $$ (x,y)^T + t (1/\sqrt{2},1/\sqrt{2}) \quad (t \in \mathbb{R}) $$ After a distance $d$ line $b$ intersects the ellipse at $(u,v)^T$ $$ 4 = 4u^2+v^2 \Rightarrow \\ v = 2 \sqrt{1-u^2} $$ where we picked the positive root for the intersection point on the ellipse closer to the line $a$. We have $$ (x, y) - d (1/\sqrt{2},1/\sqrt{2}) = (u, v) $$ We must have $$ (u + d/\sqrt{2}, 2\sqrt{1-u^2}+d/\sqrt{2}) = (u + d/\sqrt{2}, 10-u-d/\sqrt{2}) $$ where the equation for the $y$-components gives the distance in terms of $u$: $$ 2\sqrt{1-u^2}+d/\sqrt{2} = 10-u-d/\sqrt{2} \iff \\ d = \frac{10-u-2 \sqrt{1-u^2}}{\sqrt{2}} $$ We want to minimize $d$ in terms of $u$ and look at the derivative $$ d' = -\frac{1}{\sqrt{2}} + \sqrt{2} \frac{u}{\sqrt{1-u^2}} $$ For a local extrema this must vanish: $$ \frac{u}{\sqrt{1-u^2}} = \frac{1}{2} \Rightarrow \\ u = 1/\sqrt{5} \Rightarrow \\ v = 4/\sqrt{5} $$ Here the result is $(u,v) = (0.447, 1.789)$.

Comparison:

The measurement had an error of $13\%$ for $u$ and $3\%$ for $v$. The calculated minimal distance is $5.49$, thus the same as the measured one.

Answer 3

similar to Max perpendicular distance between line and parabola: find the tangent parallel to the line.

a way to construct the solution using compass and ruler can be found here

Answer 4

When $A,B$ are not both zero, the distance from a point $(u,v)$ to the line $A x +B y+C=0$ is $|A u+B v+C|/\sqrt {A^2+B^2}.\;$ For a point $(u,v)$ on the ellipse $4 x^2+y^2=4,\;$ let $u=\cos t$ and $v=2\sin t.$ The distance from $(\cos t, 2\sin t)$ to the line $x+y-10=0$ is $ \;|\cos t+2\sin t -10|/\sqrt 2.\;$ The largest value of $(\cos t+2\sin t)$ is $\sqrt 5,$ which is attained only when $\cos t=1/\sqrt 5$ and $\sin t=2/\sqrt 5.$ So the shortest distance from the ellipse to the line is $(10-\sqrt 5)/\sqrt 2\;$ which occurs only when $(u,v)=(\cos t,2\sin t)=(1/\sqrt 5,\;4/\sqrt 5).$