Question 1016364
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.
--------------
{{{ x^2/64 + y^2/25 = 1}}}
Find the slope of the ellipse at any point.
Slope m = dy/dx
{{{ x^2/64 + y^2/25 = 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/(64y) = sqrt(3)}}}
{{{x = - 64*sqrt(3)y/25}}}
{{{ x^2/64 + y^2/25 = 1}}} --> {{{25x^2 + 64y^2 = 1600}}}
Sub for x in {{{25x^2 + 64y^2 = 1600}}}
---
{{{4096*3y^2/25 + 64y^2 = 1600}}}
{{{12288y^2 + 1600y^2 = 40000}}
{{{13888y^2 = 40000}}}
{{{217y^2 = 625}}}
{{{y^2 = 625/217}}}
Sub for y^2 in {{{25x^2 + 64y^2 = 1600}}} to find x^2
x^2 = 12288/217 = 4096*3/217
{{{x = +64*sqrt(3/217)}}}
{{{x = -64*sqrt(3/217)}}}
Solve {{{25x^2 + 64y^2 = 1600}}} for y^2
{{{y^2 = 625/217}}}
{{{y = +25/sqrt(217)}}}
{{{y = -25/sqrt(217)}}}
---------------
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))