The Shadow Knows

BigGlenntheHeavy

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

shadow created by the elliptical region x2+4y2 = 5. If the point (5,0) is on\displaystyle shadow \ created \ by \ the \ elliptical \ region \ x^2+4y^2 \ = \ 5. \ If \ the \ point \ (-5,0) \ is \ on

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

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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:

5x22=x25x2(x+5)\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.

14=a8\displaystyle \frac{1}{4}=\frac{a}{8}

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

The height is 2. Is that it?.

The line has equation y=14x+54\displaystyle y=\frac{1}{4}x+\frac{5}{4}

The other line has equation y=114x254\displaystyle y=\frac{11}{4}x-\frac{25}{4}
 

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

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

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

Then m1 = yx+5\displaystyle Then \ m_1 \ = \ \frac{y}{x+5}

Now implicit differentation of x2+4y2 = 5 gives y = m2 = x4y\displaystyle Now \ implicit \ differentation \ of \ x^2+4y^2 \ = \ 5 \ gives \ y' \ = \ m_2 \ = \ \frac{-x}{4y}

Ergo, m1 = m2      yx+5 = x4y      4y2 = x25x\displaystyle Ergo, \ m_1 \ = \ m_2 \ \implies \ \frac{y}{x+5} \ = \ \frac{-x}{4y} \ \implies \ 4y^2 \ = \ -x^2-5x

     x2+4y2 = 5x, but we were given that x2+4y2 = 5, hence\displaystyle \implies \ x^2+4y^2 \ = \ -5x, \ but \ we \ were \ given \ that \ x^2+4y^2 \ = \ 5, \ hence

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

Therefore, y0 = 14(x+5)      y = x+54, equation of the line.\displaystyle Therefore, \ y-0 \ = \ \frac{1}{4}(x+5) \ \implies \ y \ = \ \frac{x+5}{4}, \ equation \ of \ the \ line.

Hence, when x = 3, y = h = 2, therefore the height of the lamp = 2 units. QED\displaystyle Hence, \ when \ x \ = \ 3, \ y \ = \ h \ = \ 2, \ therefore \ the \ height \ of \ the \ lamp \ = \ 2 \ units. \ QED
 
Hello, BigGlenn!

Here's another approach . . .


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

and a shadow created by the elliptical region, x2+4y2 = 5\displaystyle x^2+4y^2 \ = \ 5

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

The equation of the tangent to the ellipse at (x1,y1) is given by:   x1x  +  4y1y   =   5\displaystyle \text{The equation of the tangent to the ellipse at }(x_1,y_1)\text{ is given by: }\;x_1x \;+\;4y_1y \:\;=\;\:5


Since (-5,0) is on this line:   (-5)x+4(0)y=5x=1y=1\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

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


The slope of AP is:   m=10-1(-5)=14\displaystyle \text{The slope of }AP\text{ is: }\;m \:=\:\frac{1-0}{\text{-}1-(\text{-}5)} \:=\:\frac{1}{4}

The equation of the tangent is:   y0  =  14(x[-5])y=14x+54\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}

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


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

 
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