# Thread: Equations of tangent lines of an ellipse

1. ## Equations of tangent lines of an ellipse

How do I solve this question?

Find the equations of both the tangent lines to the ellipse x^2 + 4y^2 = 36 that pass through the point (12, 3).

There is supposed to be 2 tangent lines, a horizontal and non-horizontal one.

2. Originally Posted by zer0e
How do I solve this question?

Find the equations of both the tangent lines to the ellipse x2 + 4y2 = 36 that pass through the point (12, 3).

There is supposed to be 2 tangent lines, a horizontal and non-horizontal one.

$x^2+4y^2=36$

The second you hear tangent line, think derivative. Since this has a $y^2$ in it, we're going to differentiate implicity

$2x+8y\frac{dy}{dx}=0$

Moving stuff around,

$\frac{dy}{dx}=-\frac{2x}{8y}=-\frac{x}{4y}$

So the slopes of our lines must have the form above to be tangent

Now we have 1 point, so let's find the toher point

Points on the elipse are represented by $(x,\sqrt{\frac{36-x^2}{4}})$ or $(x,-\sqrt{\frac{36-x^2}{4}})$

And slope= $\frac{y_1-y_2}{x_1-x_2}$

Sooooo,

The derivative equation must equal the slope between the 2 points and lets use our formulas to plug in for y

$-\frac{x}{4\sqrt{\frac{36-x^2}{4}}}=\frac{3-\sqrt{\frac{36-x^2}{4}}}{12-x}$

And don't forget that the y values can be represented 2 ways (+ and -)

$-\frac{x}{4\cdot -\sqrt{\frac{36-x^2}{4}}}=\frac{3+\sqrt{\frac{36-x^2}{4}}}{12-x}$

Solving both of those will get you your x values, and then to get your y's remember to keep to + and -'s straight

Then the equation of the tangent lines will be

$y-y_0=-\frac{x}{4y}(x-x_0)$

Of course, you already said 1 tangent line will be horizontal, and you know the point it goes through, so you can get that equation almost without thinking

### find equations of both the tangent lines to the ellipse x 2 4y 2 36

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