# Thread: About tangent to the ellipse

1. ## [SOLVED] About tangent to the ellipse

Given the equation of the ellipse $x^2+12y^2=48$
Find the equations of the tangents drawn from the point (4,6) and thier points of contact with the ellipse.

I got the gradient of the tangent which is $\frac{\,dy}{\,dx}=-\frac{x}{12y}$
Then I equate it with the gradient of the line
$-\frac{x}{12y}=\frac{y-6}{x-4}$

$-x(x-4)=12y(y-6)$

$12y^2+x^2=72y+4x$

$72y+4x=48$ Here is when Im stuck

2. Originally Posted by imforumer
Given the equation of the ellipse $x^2+12y^2=48$
Find the equations of the tangents drawn from the point (4,6) and thier points of contact with the ellipse.

I got the gradient of the tangent which is $\frac{\,dy}{\,dx}=-\frac{x}{12y}$
Then I equate it with the gradient of the line
$-\frac{x}{12y}=\frac{y-6}{x-4}$

$-x(x-4)=12y(y-6)$

$12y^2+x^2=72y+4x$

$72y+4x=48$ Here is when Im stuck
You've done well. Now that can easily be reduced to x= 12- 18y= 6(2- 3y). Put that into the equation for the ellipse and you have a single quadratic equation to solve for y.

3. but what the straight line $72y+4x=48$ stands for?