The Shadow Knows

[BigGlenntheHeavy]

\(\displaystyle The \ figure \ shows \ a \ lamp \ located \ three \ units \ to \ the \ right \ of \ the \ y-axis \ and \ a\)

\(\displaystyle shadow \ created \ by \ the \ elliptical \ region \ x^2+4y^2 \ = \ 5. \ If \ the \ point \ (-5,0) \ is \ on\)

\(\displaystyle the \ edge \ of \ the \ shadow, \ how \ far \ above \ the \ x-axis \ is \ the \ lamp \ located?\)

[attachment=0:3ibb9tb8]AAA.JPG[/attachment:3ibb9tb8]

 

[galactus]

Let's see if I have it right, Big G.

This is just one way to go about it I am sure.

To find where the line is tangent to the ellipse:

\(\displaystyle \frac{\sqrt{5-x^{2}}}{2}=\frac{-x}{2\sqrt{5-x^{2}}}(x+5)\)

Solving for x, we get it is tangent at (-1,1)

This triangle formed is similar to the one we need the height of.

Let 'a' be the height we are looking for.

\(\displaystyle \frac{1}{4}=\frac{a}{8}\)

\(\displaystyle \boxed{a=2}\)

The height is 2. Is that it?.

The line has equation \(\displaystyle y=\frac{1}{4}x+\frac{5}{4}\)

The other line has equation \(\displaystyle y=\frac{11}{4}x-\frac{25}{4}\)

 

[BigGlenntheHeavy]

\(\displaystyle galactus, \ here \ is \ how \ I \ solve \ the \ problem.\)

\(\displaystyle Let \ l \ be \ the \ line \ going \ through \ (-5,0) \ and \ tangent \ to \ the \ ellipse \ at \ (x,y) \ and\)

\(\displaystyle passing \ through \ (3,h), \ h \ being \ the \ height \ of \ the \ lamp.\)

\(\displaystyle Then \ m_1 \ = \ \frac{y}{x+5}\)

\(\displaystyle Now \ implicit \ differentation \ of \ x^2+4y^2 \ = \ 5 \ gives \ y' \ = \ m_2 \ = \ \frac{-x}{4y}\)

\(\displaystyle Ergo, \ m_1 \ = \ m_2 \ \implies \ \frac{y}{x+5} \ = \ \frac{-x}{4y} \ \implies \ 4y^2 \ = \ -x^2-5x\)

\(\displaystyle \implies \ x^2+4y^2 \ = \ -5x, \ but \ we \ were \ given \ that \ x^2+4y^2 \ = \ 5, \ hence\)

\(\displaystyle 5 \ = \ -5x \ \implies \ x \ = \ -1, \ ergo \ y \ = \ 1 \ (y \ above \ the \ x-axis), \gives \ m \ = \ \frac{1}{4}.\)

\(\displaystyle Therefore, \ y-0 \ = \ \frac{1}{4}(x+5) \ \implies \ y \ = \ \frac{x+5}{4}, \ equation \ of \ the \ line.\)

\(\displaystyle Hence, \ when \ x \ = \ 3, \ y \ = \ h \ = \ 2, \ therefore \ the \ height \ of \ the \ lamp \ = \ 2 \ units. \ QED\)

 

[soroban]

Hello, BigGlenn!

Here's another approach . . .

The figure shows a lamp located 3 units to the right of the \(\displaystyle y\)-axis

and a shadow created by the elliptical region, \(\displaystyle x^2+4y^2 \ = \ 5\)

If the point \(\displaystyle A(-5,0)\) is the edge of the shadow, how far above the \(\displaystyle x\)-axis is the lamp located?

\(\displaystyle \text{The equation of the tangent to the ellipse at }(x_1,y_1)\text{ is given by: }\;x_1x \;+\;4y_1y \:\;=\;\:5\)


\(\displaystyle \text{Since }(\text{-}5,0)\text{ is on this line: }\;(\text{-}5)x + 4(0)y \:=\:5 \quad\Rightarrow\quad x \:=\:-1 \quad\Rightarrow\quad y \:=\:1\)

\(\displaystyle \text{The line is tangent at }P(-1,1)\)


\(\displaystyle \text{The slope of }AP\text{ is: }\;m \:=\:\frac{1-0}{\text{-}1-(\text{-}5)} \:=\:\frac{1}{4}\)

\(\displaystyle \text{The equation of the tangent is: }\;y - 0 \;=\; \tfrac{1}{4}(x-[\text{-}5]) \quad\Rightarrow\quad y \:=\:\tfrac{1}{4}x + \tfrac{5}{4}\)

\(\displaystyle \text{When }x = 3: }\;y \:=\:\tfrac{1}{4}(3) + \tfrac{5}{4} \quad\Rightarrow\quad y \:=\:2\)


\(\displaystyle \text{Therefore, the lamp is 2 units above the }x\text{-axis.}\)

 

[galactus]

3 different methods for the same problem. Cool.