watchmath
  • watchmath
Find the height of the lamp post (see the attachment).
Mathematics
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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watchmath
  • watchmath
welcome to answer everybody :).
1 Attachment
myininaya
  • myininaya
did you make this problem up or did you find it somewhere
watchmath
  • watchmath
It is from a book :). We give this as an assignment for our student :).

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myininaya
  • myininaya
i think i'm fixing to figure out lol
myininaya
  • myininaya
if i can figure out P i'm done
myininaya
  • myininaya
so far i have y1=2x0/y0 where P=(-x0,y0) this is the way i'm suppose to go about it right?
myininaya
  • myininaya
and where L representing the point of the bulb of the lamp post at (3,y1)
myininaya
  • myininaya
so found the slope of the line containing P and the x intercept (-5,0) and got m=(y1-0)/(3+5)=y1/8 and i found the derivative of the ellipse which is y'=-x/4y i evaluated y' at (-x0,y0) and got y'=x0/4y0 so this has to be equal to y1/8 so we have y1=2x0/y0
myininaya
  • myininaya
omg omg P=(-1,1) so y1=2
myininaya
  • myininaya
1 Attachment
myininaya
  • myininaya
the height is 2 :)
myininaya
  • myininaya
ty so much for the problem :) i don't know if i went about the long way or not but i feel good about it
watchmath
  • watchmath
Good job Miyaniya :) So you like it?
myininaya
  • myininaya
yes im going to use it wow it took while for me to figure how to get P lol that was my problem for the longest time
myininaya
  • myininaya
is this way that you were thinking about finding the height? or is there a shorter way?
watchmath
  • watchmath
Here what I will do to find \(P\). Let \(P=(a,b)\). Using \((a,b)\) and \((-5,0)\) the slope is \(m=\frac{b}{a+5}\). Using implicit differentiation the slope is \(m=-\frac{a}{4b}\). Setting the two to be equal, we have \(4b^2=-a^2-5a\). But since \((a,b)\) is on the ellipse, \(4b^2=5-a^2\). So \(-a^2-5a=-a^2+5\) which implies \(a=-1\). It follows that \(4b^2=5-a^2=4\). So \(b=\pm 1\). But \(b\) is above the \(x\)-axis. So \(b=1\). Hence \(P=(-1,1)\). Let \(h\) be the height that we are looking for. Using similar triangle we have \(\frac{h}{5+3}=m=\frac{1}{4}\). Then \(h=2\).
myininaya
  • myininaya
ok nice. you have less text than i have writing lol
myininaya
  • myininaya
thank you so much watchmath. i'm totally going to give this to my calculus students and let them work on it as a group project
watchmath
  • watchmath
My pleasure :)

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