SOLUTION: The ellipse {{{ x^2/64 + y^2/25 = 1}}} is given. Find the equations of the lines tangent to this ellipse which make an angle of 60 degrees with the x-axis. can someone sho

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The ellipse {{{ x^2/64 + y^2/25 = 1}}} is given. Find the equations of the lines tangent to this ellipse which make an angle of 60 degrees with the x-axis. can someone sho      Log On


   

Question 1016364: The ellipse +x%5E2%2F64+%2B+y%5E2%2F25+=+1 is given. Find the equations of the lines tangent to this ellipse which make an angle of 60 degrees with the x-axis.



can someone show me how to do and answer it. thank you very much for help
Found 3 solutions by rothauserc, robertb, Alan3354:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
The tangent of 60 degrees gives us slope of the line
:
tangent 60 = 1.732050808 approx 1.7
:
the equation of the 60 degree line passing through the origin is
y = 1.7x
:
our ellipse is centered around the origin and
x-axis intercepts are (8,0) and (-8,0)
y-axis intercepts are (0,5) and (0,-5)
:
we need to translate our line 8 units to the left and 8 units to the right
and adjust the slope sign, therefore the equations are
:
y = -1.7(x + 8)
y = 1.7(x - 8)
:

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
If the line makes an angle of 60 degrees with the x-axis, then one pair of lines would be y+=+tan%2860%5Eo%29%2Ax+%2B+k, or y+=+sqrt%283%29x%2Bk where k would have two values each of which is the negative of the other.
Substitute this into +x%5E2%2F64+%2B+y%5E2%2F25+=+1, from which we get
x%5E2%2F64+%2B+%28sqrt%283%29%2Ax%2Bk%29%5E2%2F25+=+1.
==>
x%5E2%2F64+%2B+%283x%5E2%2B2sqrt%283%29%2Akx+%2Bk%5E2%29%2F25+=+1.
After simplifying, we get
%28217%2F1600%29%2Ax%5E2+%2B+%28%282sqrt%283%29%29%2F25%29%2Akx+%2B+k%5E2%2F25-1+=+0.
Now since the line is supposed to be tangent to the ellipse, the discriminant of the quadratic equation above must be 0.
Hence
%2812%2F625%29%2Ak%5E2+-+4%2A%28217%2F1600%29%2A%28k%5E2%2F25+-1%29+=+0, or
%28-1%2F400%29%2Ak%5E2+%2B+217%2F400+=+0, or k%5E2+=+217, or k+=+sqrt%28217%29 or k+=+-sqrt%28217%29
Thus the two lines are y+=+sqrt%283%29x+%2B+sqrt%28217%29 and y+=+sqrt%283%29x+-+sqrt%28217%29.
Note that we cannot accept the other pair of lines y+=+-sqrt%283%29x+%2B+sqrt%28217%29 and y+=+-sqrt%283%29x+-+sqrt%28217%29 because these two lines make an angle of 120 degrees with the x-axis.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The ellipse +x%5E2%2F64+%2B+y%5E2%2F25+=+1 is given. Find the equations of the lines tangent to this ellipse which make an angle of 60 degrees with the x-axis.
--------------
+x%5E2%2F64+%2B+y%5E2%2F25+=+1
Find the slope of the ellipse at any point.
Slope m = dy/dx
+x%5E2%2F64+%2B+y%5E2%2F25+=+1
Differentiate implicitly
(x/32) dx + (2y/25)dy = 0
dy/dx = -25x/(64y)
-----
The slope of 60 degs with the x-axis is tan(60) = sqrt(3)
---
--> -25x%2F%2864y%29+=+sqrt%283%29
x+=+-+64%2Asqrt%283%29y%2F25
+x%5E2%2F64+%2B+y%5E2%2F25+=+1 --> 25x%5E2+%2B+64y%5E2+=+1600
Sub for x in 25x%5E2+%2B+64y%5E2+=+1600
---
4096%2A3y%5E2%2F25+%2B+64y%5E2+=+1600
12288y%5E2+%2B+1600y%5E2+=+40000%7D%7D%0D%0A%7B%7B%7B13888y%5E2+=+40000
217y%5E2+=+625
y%5E2+=+625%2F217
Sub for y^2 in 25x%5E2+%2B+64y%5E2+=+1600 to find x^2
x^2 = 12288/217 = 4096*3/217
x+=+%2B64%2Asqrt%283%2F217%29
x+=+-64%2Asqrt%283%2F217%29
Solve 25x%5E2+%2B+64y%5E2+=+1600 for y^2
y%5E2+=+625%2F217
y+=+%2B25%2Fsqrt%28217%29
y+=+-25%2Fsqrt%28217%29
---------------
For +60 degs with the x-axis, when x is +, y is negative, and vice versa.
The tangent points are:
(+64*sqrt(3/217),-25/sqrt(217))
--> y + (25/sqrt(217)) = sqrt(3)*(x - 64*sqrt(3/217))
and
(-64*sqrt(3/217),+25/sqrt(217))
--> y - (25/sqrt(217)) = sqrt(3)*(x + 64*sqrt(3/217))