CPhill

3. Find the foci of the ellipse whose major axis has endpoints (0,0) and (13,0) and whose minor axis has length 12.

We have this equation

(x - h)^2            (y - k)^2

_______    +   ________   =      1

   a^2                  b^2

The center of the ellipse is   at  (6.5, 0)     =  ( h, k)

The major axis is along the x axis.....its length is 13

"a" is 1/2 of this 6.5 

The minor axis is along y

And b is 1/2 of this  =  6

The y coordinate of both foci lies on the x axis

And the foci are given by   (  ±√ [a^2 - b^2 ] + h , 0 )    =   ( ±√ [6.5^2 - 6^2] + 6.5 , 0)   =

( 2.5 + 6.5, 0 )     and   (-2.5 + 6.5)    =

(9, 0)    and  ( 4, 0)

Here's the graph : https://www.desmos.com/calculator/hxa6eulntu

Thanks, Guest!!!

THAT is a REALLY nice method  !!!!

We have a SSA-type problem

Let the lighthouse be located at C......angle ACB  = 33°

     C

                36

A        20        B

We are looking for angle CBA

We can find angle CAB, first

Using the Law of Sines

sin ACB /20 = sinCAB / 36

sin (33) / 20  = sin(CAB) /36

(36/20)sin (33)  = sinCAB

arcsin [ (36/20)sin(33)] = CAB  ≈ 78.62°

Therefore  a possible value for  angle CBA   ≈ 180 - 33 - 78.62  ≈ 68.38°

Another possible value for angle CAB  =  180 - 78.62  ≈ 101.38°

Add  this to the known angle ACB  =  33 + 101.38   = 134.38°

Because this is < 180, we have two possible triangles

So....the other possible value for angle CBA  = 180 - 134.38  = 45.62°

This area will form a  right riangle

We can find one leg as follows

6  = (1/2)x     multiply both sides by 2

12  = x

The other leg  = 6

And the area of a right triangle =    product of the legs / 2   =

[ 12 * 6 ] /   2      =

72  / 2  =

36 units^2  

Here's a graph : https://www.desmos.com/calculator/0vjigygyh7

Thanks to X², Melody and Alan.....!!!!!

I agree with X².....this one is difficult.....!!!

I took an alternative approach....in the interest of space cosiderations, I will use some tecnology to help with the answer....but....I get the same answer as X² did!!!

First of all..... I wanted to find the point  where the tangent line intersects the x axis

Let this point =  A

Call  the distance between (2, 0) and A =  x

So..  using similar triangles..we have this relationship

[ x ] / [ 6 + x]  =  1 / 9

9x  =  6 + x

8x  = 6

x = 6/8  =  3/4

So....the point is  ( 2 - x , 0)  =  ( 2 - 3/4, 0)  =  (1.25, 0)  = A

Now we can find the distance betwwen the center of the smaller circle and A by using similar triangles again....call the distance, y....so we have

[ 1 + y] / [ 11 + y]  = [ 1 ] / [ 9]

9 [ 1 + y ]  = 11 + y

9 + 9y  = 11 + y

8y = 2

y = 1/4   = .25

And we can find the distance  between A  and the point where the tangent line touches the smaller circle  as    sqrt [ (1 + 1/4)^2  - 1]  =  sqrt [ (5/4)^2 - 1]  = sqrt [ (25 -16)/ 16] = sqrt (9/16) =

3/4   = .75

And we can find the point where the tangent line touches the smaller circle by using a little technology to find the intersection of these two circles :

(x - 2)^2  + ( y - 1)^2  = 1

(x - 1.25)^2 + y^2  = .75^2      

(1.04, 0.72)  =  B

Likewise......we can find the distance  between A  and the point where the tangent line touches the larger circle  as    sqrt  [ 11.25^2   - 9^2]  =  6.75

And........We can find the point where the tangent line touches the larger circle by using a little technology to find the intersection of these two circles :

(x - 8)^2  + (y - 9)^2  =  81

(x - 1.25)^2 + y^2  = 6.75^2    

( -0.64, 6.48) = C

The slope between B and C  is

[ 6.48 - 0.72] / [ -0.64 - 1.04)  =  [ 5.76 ] / [-1.68]  = -24/7

So....using  A  and this slope we can write an equation of the tangent line as

y  = (-24/7)(x - 1.25)

y = (-24/7) (x - 5/4)

y = (-24/7)x + 120/ 28

y = (-24/7)x + 30/ 7

So...."b"   =  30/7

HAHAHA!!!!!.....that sounds better, X²......!!!!!

Glad to help  !!!

Thanks....sometimes it helps to  "un-rationalize" the denominator on those square root ratios   ....[if that's even a word.....LOL!!!!  ]

Note    cos (-θ)   =  cos(θ)

So

cos(θ)  = √3/4  =  x / r

And

sin (θ)  =  y / r

We need to find "y"  ....since  r is always positive  and the sine < 0....then y  must be negative

So

y   =  -sqrt  ( r^2  - x^2) =  - sqrt ( 4^2 - (sqrt(3) )^2 )  =  - sqrt ( 16 - 3)  =  -sqrt (13)

So.....

sin (θ)  =  - √13  / 4    ⇒   C