# Tangent to Ellipse

1. Feb 20, 2013

### crowdedscience

An ellipse has the equation x^2+5y^2=5
a line has the equation y=mx+c
a) show that if the line is a tangent to the ellipse then c^2=5m^2+1
b) hence find the equation of the tangent parallel to the line x-2y+1=0

I tried to find the gradient of x^2+5y^2=5 at a point (x1,y1) and then put it into the equation (y-y1)=m(x-x1), but that didnt work out
any ideas?

2. Feb 20, 2013

### Simon Bridge

If the line is a tangent to the ellipse, then it intersects at only one point.

3. Feb 20, 2013

### HallsofIvy

Staff Emeritus
Simon Bridge's point is that y= mx+ c intersects $x^2+ 5y^2= 5$ then the quadratic equation $x^2+ 5(mx+ c)^2= 5$ has at least one solution. The line is tangent to the ellipse if and only if that quadratic equation has a single solution- its discriminant is 0. That's a method, due to Fermat, that predates Calculus.

But you certainly can do this by finding the gradient of y. What did you get for the gradient?

4. Feb 20, 2013

### crowdedscience

the derivative of x^2+5y^2=5 was dy/dx=-x/5y

5. Feb 21, 2013

### Simon Bridge

... so, if (p,q) is a point on the ellipse, then the gradient of the tangent at that point is
m=-p/5q and the equations of the tangent line is y=mx+c ... so how do you find c?

I didn't know the other method was from Fermat :)

6. Feb 21, 2013

### crowdedscience

ok, so I did
y-y1=m(x-x1)
y-q=-p/5q(x-p)
y=-px/5q + p^2/5q + q
y=(-p/5q)x + (p^2/5q + q)
so then c=p^2/5q + q

but then when I put that into c^2=5m^2+1, it isn't equal
what did I do wrong?

Thanks

7. Feb 21, 2013

### Simon Bridge

note: c=p^2/5q + q
means: c=q(1+5m2) ... encouraging?
but what you are looking for is c2.

I, personally, wouldn't have tried it this way.
If you use Fermat's approach, the relation just drops out.

8. Feb 21, 2013

### crowdedscience

sorry, I'm starting to feel stupid now, but I just don't see how (q(1+5m^2))^2=5m^2+1. Also I'm not familiar with fermat's approach, what does that involve?

9. Feb 21, 2013

### Simon Bridge

It doesn't. You wouldn't expect it to because that would mean that c=c^2. It is quite close though.
You just need to figure out how to get from c=q(1+5m^2) to c^2=1+5m^2 ... looks like we are both missing something. Like I said, I wouldn't normally do it this way.
Hopefully HallsofIvy will pop back in to steer us right ;)
See posts #2 and #3.

10. Feb 21, 2013

### HallsofIvy

Staff Emeritus
Since you ask (hope I don't get into trouble for this!)

Fermat's method would be this: the ellipse, $x^2+ 5y^2= 5$ and the line y= mx+ c will intersect when $x^2+ 5(mx+ c)^2= x^2+ 5m^2x^2+ 10mcx+ 5c^2= 5$ or $(1+ 5m^2)x^2+ 10mcx+ 5(c^2- 1)= 0$, a quadratic equation for x. They will be tangent when that has a double root- when its discriminant is 0.

The discriminant is
$$100m^2c^2- 4(1+ 5m^2)5(c^2- 1)= 100m^2c^23- 20(c^2- 1+5m^2c^2- 5m^2)= 0$$
Can you simplify that?

11. Feb 21, 2013

### Simon Bridge

@HallsofIvy: I had hoped you show the next step for the differential method...