Question

Find the height of the lamp post (see the attachment).

Answer 1

welcome to answer everybody :).

Answer 2

did you make this problem up or did you find it somewhere

Answer 3

It is from a book :). We give this as an assignment for our student :).

Answer 4

i think i'm fixing to figure out lol

Answer 5

if i can figure out P i'm done

Answer 6

so far i have y1=2x0/y0 where P=(-x0,y0)
this is the way i'm suppose to go about it right?

Answer 7

and where L representing the point of the bulb of the lamp post at (3,y1)

Answer 8

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

Answer 9

omg omg P=(-1,1) so y1=2

Answer 10

the height is 2 :)

Answer 11

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

Answer 12

Good job Miyaniya :)
So you like it?

Answer 13

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

Answer 14

is this way that you were thinking about finding the height?
or is there a shorter way?

Answer 15

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 \(h=2\).

Answer 16

ok nice. you have less text than i have writing lol

Answer 17

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

Answer 18

My pleasure :)