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.
Thank you in advance!
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.
Thank you in advance!
$\displaystyle x^2+4y^2=36$
The second you hear tangent line, think derivative. Since this has a $\displaystyle y^2$ in it, we're going to differentiate implicity
$\displaystyle 2x+8y\frac{dy}{dx}=0$
Moving stuff around,
$\displaystyle \frac{dy}{dx}=-\frac{2x}{8y}=-\frac{x}{4y}$
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 $\displaystyle (x,\sqrt{\frac{36-x^2}{4}})$ or $\displaystyle (x,-\sqrt{\frac{36-x^2}{4}})$
And slope=$\displaystyle \frac{y_1-y_2}{x_1-x_2}$
Sooooo,
The derivative equation must equal the slope between the 2 points and lets use our formulas to plug in for y
$\displaystyle -\frac{x}{4\sqrt{\frac{36-x^2}{4}}}=\frac{3-\sqrt{\frac{36-x^2}{4}}}{12-x}$
And don't forget that the y values can be represented 2 ways (+ and -)
$\displaystyle -\frac{x}{4\cdot -\sqrt{\frac{36-x^2}{4}}}=\frac{3+\sqrt{\frac{36-x^2}{4}}}{12-x}$
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
$\displaystyle y-y_0=-\frac{x}{4y}(x-x_0)$
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