Conics question (1 Viewer)

[jkwii]

hey how do u find the equations of the tangents to an ellipse from an external point?

 

[nottellingu]

Okay ill take a wild guess. Find the equation of the chord of contact from that external point. Then solve simultaneously to find the points where the chord of contact intersects with the ellipse. Then u can go about normally by finding the tangent to the ellipse at these points

 

[jkwii]

thanks so much!

that rally helped a lot

 

[kony]

that method depends on whether the chord of contact formula can be used or not. If not, it is much longer than the discriminants method.

 

[ssglain]

that method depends on whether the chord of contact formula can be used or not. If not, it is much longer than the discriminants method.

But it only takes about half a dozen lines to prove the chord of contact formula!

Somehow the discriminant method doesn't sound very pleasant to me in this case. Let's just think about what we need to do before we do it. This is a fairly standard procedure, I think:
1. Take a random linear equation y = mx + c which passes through some point (j, k) external to the ellipse. So k = mj + c --> c = k - mj, and the line becomes y = mx + (k - mj).
2. Intersect y = mx + (k - mj) with the ellipse and solve simultaneously for the x-coordinate of the point of intersection. This will produce a quadratic equation in x.
3. Force discriminant = 0 for tangential behaviour and find the value of m when this happens.

It just seems that this will turn out to be a huge algebraic crunch that I am personally inclined to avoid.

So please do enlighten me if I've taken this on from the heavier side.

 

[samwell]

find the general gradient of the point of intersection which must be:
2x/a^2+dy/dx(2y/b^2)=o
dy/dx=-x(b^2)/y(a^2)
if (x1,y1) is the xternal pt
(y-y1)=-x(b^2)/y(a^2)[x-x1]
x^2/a^2+y^2/b^2=x.x1/a^2+y.y1/b^2 [sub in the x1,y1]
Although this will onli give u one eqn.