Solving Tangent to Ellipse: x^2+5y^2=5, y=mx+c

In summary: OP is using that as a model.Sure, no problem. So, we have the derivative dy/dx=-x/5y. Now, we know the equation of the tangent line is y=mx+c. So, we can plug in the coordinates of the point of tangency (p,q) to get q=mp+c. Then, we can substitute this into the derivative to get -p/5q=m. Solving for c, we get c=q(1+5m^2). Now, we can substitute this into c^2=5m^2+1 to get q^2(1+5m^2)^2=5m^2+1. Simplifying
  • #1
crowdedscience
6
0
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?
 
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  • #2
If the line is a tangent to the ellipse, then it intersects at only one point.
 
  • #3
Simon Bridge's point is that y= mx+ c intersects [itex]x^2+ 5y^2= 5[/itex] then the quadratic equation [itex]x^2+ 5(mx+ c)^2= 5[/itex] 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
the derivative of x^2+5y^2=5 was dy/dx=-x/5y
 
  • #5
... 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
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
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
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
I just don't see how (q(1+5m^2))^2=5m^2+1.
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 into steer us right ;)
Also I'm not familiar with fermat's approach, what does that involve?
Quadratic equation.
See posts #2 and #3.
 
  • #10
Since you ask (hope I don't get into trouble for this!)

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

The discriminant is
[tex]100m^2c^2- 4(1+ 5m^2)5(c^2- 1)= 100m^2c^23- 20(c^2- 1+5m^2c^2- 5m^2)= 0[/tex]
Can you simplify that?
 
  • #11
@HallsofIvy: I had hoped you show the next step for the differential method...
 

1. What is the formula for finding the tangent to an ellipse?

The formula for finding the tangent to an ellipse is y=mx+c, where m is the slope of the tangent line and c is the y-intercept.

2. How do you solve for the slope of the tangent line?

To solve for the slope of the tangent line, you can use the derivative of the ellipse equation (x^2+5y^2=5) and solve for y'. The resulting equation will give you the slope of the tangent line at any given point on the ellipse.

3. What is the significance of the y-intercept in the equation of the tangent line?

The y-intercept (c) represents the point where the tangent line intersects the y-axis. This point can be used to plot the tangent line on a graph and determine the location of the tangent point on the ellipse.

4. Can you solve for the tangent line at any point on the ellipse?

Yes, you can solve for the tangent line at any point on the ellipse by plugging in the x and y coordinates of the point into the equation y=mx+c. This will give you the equation of the tangent line at that specific point.

5. How do you graph the tangent line on an ellipse?

To graph the tangent line on an ellipse, you can plot the y-intercept (c) on the y-axis and use the slope (m) to plot another point on the tangent line. You can then connect these two points to create the tangent line on the ellipse.

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