Problem involving implicit differentiation over an ellipse

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The discussion revolves around solving a calculus problem involving implicit differentiation of an ellipse defined by the equation x^2 + 4y^2 = 5. The challenge is to determine the height of a lamppost located at x=3, which casts a shadow that touches the ellipse at a specific point. Participants suggest using parametric equations for the ellipse and finding the intersection with the shadow line defined by y = (x + 5)h / 8, where h is the height of the lamp. The key to solving the problem lies in setting up a quadratic equation from the intersection and ensuring it has one solution by making the discriminant zero. This approach allows for the calculation of the lamp's height based on the geometry of the situation.
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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