2x^2 + y^2 = 54 from (10,1)
P.S. I still don't study calculus. This lesson is from analytic geometry and I have no idea how to solve it because my professor didn't teach it. So if someone could tell me step-by-step on how to do it , I would appreciate it very much.
I really need it later so please don't put the question on hold again. Thanks.
To follow through on André Nicholas' comment (I'd proposed this approach, but dropped the comment when the question looked like it would get "closed"), the equation for the ellipse becomes
$$ 2X^2 \ + \ (mX + [1 - 10m])^2 \ = \ 54 \ , $$
where $ \ (X,Y) \ = \ (X, mX+[1−10m] ) \ \ $ is the tangent point on the ellipse. It is important in making the computation at this point not to multiply everything out (the other reason I dropped my comment, since the actual execution of this approach has to be carried out carefully to avoid insanity, and needs more than a comment to discuss). The resulting quadratic equation in $ \ X \ $ should best be written
$$ (m^2 + 2) \ X^2 \ + \ 2m \ (1 - 10m) \ X \ + \ [ \ (1-10m)^2 - 54 ] \ = \ 0 , $$
for which the discriminant is
$$ \ D \ = \ 4m^2 \ (1-10m)^2 \ - \ 4 \cdot (m^2 + 2) \cdot [ \ (1-10m)^2 - 54 ] \ \ . $$
In order to have a tangent point, the line must intersect the ellipse for a single value of $ \ X \ , $ so we want this discriminant to be zero:
$$ 4m^2 \ (1-10m)^2 \ - \ 4 \cdot (m^2 + 2) \cdot [ \ (1-10m)^2 - 54 ] \ = 0 $$
$$ \Rightarrow \ \ m^2 \ (1-10m)^2 \ = \ (m^2 + 2) \cdot [ \ (1-10m)^2 - 54 ] $$
$$ \Rightarrow \ \ 0 \ = \ 2 \cdot (1-10m)^2 \ - \ (m^2 + 2) \cdot 54 $$
$$ \Rightarrow \ \ 146m^2 \ - \ 40m^2 \ - \ 106 \ = 0 \ \ . $$
In this way, we get the quadratic equation to solve for the slopes of the tangent lines, rather than a rather scary quartic equation. Here is a graph of the geometrical arrangement.
When you reach calculus, you'll learn a much simpler way to deal with this sort of problem (and one applicable to curves more complicated than conic sections).