Ellipse Question (1 Viewer)

cutemouse · Mar 22, 2009

Hello,

Could someone please help me with the following question?

Find the equation of the tangents to the ellipse 4x²+9y²=36 given that these tangents are perpendicular to the line 3x+2y=5

I can do it using parametrics but I can't seem to do it the other way, where you make a new equation of a line with the gradient of the tangent and a constant (eg. k). Could someone please show me using that method?

Thanks

Iruka · Mar 22, 2009

OK, a line that is perpendicular to the given line is going to have the form
2x-3y= k for some constant k.

Rearrange this as 2x=3y+k, and substitute this into the equation for the ellipse. You should get a quadratic in y. Find the discriminant of the quadratic in terms of k and find conditions on k that make the discriminant= 0.

jet · Mar 22, 2009

$\text{ Well, the gradient of the line is } m = \frac{-3}{2} \\ \therefore m_{Tangent} = 2/3 \\ y - y_1 = \frac{2}{3}(x - x_1) \\ 3y +2x_1 - 3y_1 = 2x \\ x = \frac{3y + k}{2} \\ \\ \therefore 4\left ( \frac{3y + k}{2} \right )^2 + 9y^2 = 36 \\ (9y^2 + 6ky + k^2) + 9y^2 - 36 = 0 \\ 18y^2 + 6ky + (k^2 - 36) = 0 \\ \text{Now, for a tangent, } \Delta = 0 \\ 36k^2 - 4(18)(k^2 - 36) = 0 \\ 36k^2 - 72k^2 + 2592 = 0 \\ 36k^2 = 2592 \\ k^2 = 72 \\ k = \pm 6\sqrt{2}$
I checked it, and it works on a graphing program, so Im pretty sure its right.

Ellipse Question (1 Viewer)

cutemouse

Account Closed

Iruka

Member

jet

Banned

Users Who Are Viewing This Thread (Users: 0, Guests: 1)