# Thread: tangent of an ellipse

1. ## tangent of an ellipse

x^2 + 2x + 4y^2 - 16y + 13v = 0 represents an ellipse

the line y=mx meets the ellipse

show that 3m^2 - 4m - 3 => 0.

What value(s) of m make the line a tangent to the ellipse?

2. Originally Posted by deej813
x^2 + 2x + 4y^2 - 16y + 13v = 0 represents an ellipse

the line y=mx meets the ellipse

show that 3m^2 - 4m - 3 => 0.

What value(s) of m make the line a tangent to the ellipse?
1. I assume that you mean:

$x^2 + 2x + 4y^2 - 16y + 13 = 0$

2. Calculate the points of intersection between the line and the ellipse substituting y = mx:

$x^2 + 2x + 4m^2 x^2 - 16 m x + 13 = 0$ and solve for x. (Use the formula to solve a quadratice equation)

$x = - \dfrac{2\cdot \sqrt{3m^2 - 4m - 3} - 8m + 1}{4m^2 + 1} \vee x = \dfrac{2\cdot \sqrt{3m^2 - 4m - 3} + 8m - 1}{4m^2 + 1}$

3. You only get one point of intersection = tangent point if the radicand equals zero.

4. To calculate the values of m, solve for m:

$3m^2 - 4m - 3 = 0$ (Use the formula to solve a quadratice equation)

3. ok sorry still confused
what are the steps for how you get the equations in 2. the x=-... and x=...
why are there two??
how do you solve it for m?

4. Originally Posted by deej813
ok sorry still confused
what are the steps for how you get the equations in 2. the x=-... and x=...
why are there two??
how do you solve it for m?
$x^2 + 2x + 4m^2 x^2 - 16 m x + 13 = 0\Leftrightarrow (1+4m^2)x^2+(2-16m)x+13$.Then solve it as a quadratic equation.

5. um ok i tried using the quadratic formula and got to

[-2(1-8m)+-sqrt(16(3m^2-4m-3))] / 52(1-8m)

is that right?
what do i do from here??

6. Originally Posted by deej813
um ok i tried using the quadratic formula and got to

[-2(1-8m)+-sqrt(16(3m^2-4m-3))] / 52(1-8m)

is that right?
what do i do from here??
No, it should be the same answer as earboths'. Then you can prove the first part since a negative number have no square.
If y=mx is a tangent, it means it has only one intersection with the ellipse. So sqrt(...) must be zero!