Expert: Alon Mandes - 11/5/2009

Question
Find the equations of both tangent lines to the ellipse ((x^2)/4)+((y^2)/9)=1 that pass through the point (4,0).

Answer
Let's suppose that the point where the tangent line touch the ellipse is (xo,yo). We can find
a relation between xo %26 yo, by substituting (xo,yo) in the equation of the ellipse :
(xoČ/4)+(yoČ/9)=1 --> (yoČ/9)=1-(xoČ/4) --> yoČ=9[1-(xoČ/4)] --> yo=±3√[1-(xoČ/4)].
Let's call this Eq#1 .
The slope of the tangent line is equal to the derivative of the curve at point(xo,yo). Let's
derive : (2xo/4)+(2yo/9)y'=0 --> y'=-9xo/4yo=m . Let's call it Eq#2 . By substituting Eq#1 we gain : m=-9xo/{4*±3√[1-(xoČ/4)]} --> m=±xo/{4√[1-(xoČ/4)]} . Let's call this Eq#3.
The equation of the tangent line is : Y=mX+n . We can find a relation between m %26 n , by
substituting the point(4,0) in the tangent line equation : 0=4m+n . Let's call this Eq#4 .
The tangent line passes also through the point (xo,yo), therefore : yo=mxo+n. Let's call
this equation #5 . So, we have 5 equations with 4 variables. Let's list them :
#1 : yo=±3√[1-(xoČ/4)]
#2 : m=-9xo/4yo --> yo=-9xo/m
#3 : m=±xo/{4√[1-(xoČ/4)]}
#4 : 0=4m+n --> n=-4m
#5 : yo=mxo+n

I will leave it to from here to continue ..

Alon.