Problem involving implicit differentiation over an ellipse

hexag1
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Homework Statement


So here's a question from my textbook 'Calculus: Concepts and Contexts' 2nd ed. by James Stewart. This is section 3.6 # 54

We have Cartesian coordinates set up with an ellipse at x^2 + 4y^2 = 5

To the right of the ellipse a lamppost (in 2D!) stands at x=3 with unknown height. The lamppost shines a light to the left over the ellipse. The ellipse then casts a shadow. The point at (-5,0) marks where the edge of the shadow crosses the x-axis. The shadow-line is a line tangential to the ellipse running from the lamplight to (-5,0). This is the only given value for the shadow-line. The shadow-line touches the ellipse on the top left quadrant.

The Question: how tall is the lamp?

Implicit differention with respect to x gives:

2x + 8y*y' = 0

solving for y' we have: y' = -x/4y

Homework Equations



ellipse : x^2 + 4y^2 = 5

derivative of ellipse : 2x + 8y*y' = 0

shadow-line intercept at (-5,0)

The Attempt at a Solution



I find it difficult to see how to proceed. I can find expressions for various elements of the problem, but they all seem to be written in terms of each other with no way to find a number for the height of the lamp.
If I call the point where the shadow line (which is tangential to the ellipse) intercepts the ellipse (j,k) then I find that the height of the lamp is -2j/k
 
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welcome to pf!

hihexag1! welcome to pf! :smile:

one way is to call a general point on the ellipse x = 5cosθ, y= 2.5 sinθ :wink:
 
My solution would be to define the shadow line by:
y=(x+5) h / 8
where h is the height of the lamp.

Intersect with the ellipse and find out for which h it will have only 1 solution for x.

Would this yield an acceptable solution or do you have to use derivatives?
 
I like Serena said:
My solution would be to define the shadow line by:
y=(x+5) h / 8
where h is the height of the lamp.

Intersect with the ellipse and find out for which h it will have only 1 solution for x.

Would this yield an acceptable solution or do you have to use derivatives?

I don't HAVE to do derivatives, but the problem is taken from the section on implicit differention, so I assumed that was the way to go about it.

I have tried intersecting the line and the ellipse, but I can only find equations for the line where the slip and 'b' (of y=mx+b format) are written in terms of the intersecting points, which I am trying to find in the first place! I find an equation for the line, but any and all substitutions that I make to and from the ellipse, I end up going in circles with no way to get a hold of the coordinates of the intersection.
I can make an estimate of the intersection point, of course - its y-value is a bit less than
(5/4)^(1/2) <--sorry I'm having LaTex problems in chrome.
 


tiny-tim said:
hihexag1! welcome to pf! :smile:

one way is to call a general point on the ellipse x = 5cosθ, y= 2.5 sinθ :wink:

Right - so I could write the ellipse as a squashed circle. But how do I find Θ ??
 
first find the slope of the tangent at a general point θ, and use that to find the equation of the tangent :smile:

(alternatively, as you say, it's a squashed circle :biggrin:

so unsquash it by changing the coordinates, and then use simple trig!)

of course, I like Serena's :smile: method is also very simple, involving no calculus, and nothing more tricky than complete the square :wink:
 
hexag1 said:
I don't HAVE to do derivatives, but the problem is taken from the section on implicit differention, so I assumed that was the way to go about it.

I have tried intersecting the line and the ellipse, but I can only find equations for the line where the slip and 'b' (of y=mx+b format) are written in terms of the intersecting points, which I am trying to find in the first place! I find an equation for the line, but any and all substitutions that I make to and from the ellipse, I end up going in circles with no way to get a hold of the coordinates of the intersection.
I can make an estimate of the intersection point, of course - its y-value is a bit less than
(5/4)^(1/2) <--sorry I'm having LaTex problems in chrome.

If you substitute
y=(x+5) h / 8
into
x²+4y²=5

you'll get a quadratic equation of the form
ax²+bx+c=0

It has 1 solution if the discriminant is zero:
D=b²-4ac=0

Solve h from this equation.
 
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