Given an the equation of an ellipse, say $\displaystyle \frac{x^2}{9}+\frac{y^2}{16}=1$,
find the equations of the two lines tangent to this ellipse passing through the external point
P(5,6). Thank you in advance.

Originally Posted by Kaloda

Given an the equation of an ellipse, say $\displaystyle \frac{x^2}{9}+\frac{y^2}{16}=1$,
find the equations of the two lines tangent to this ellipse passing through the external point
P(5,6). Thank you in advance.

1. Solve the equation for y^2
2. Implicit differentiation

That gives the equation for the tangent lines.
3. Plug in point P.
4. What can you come up with?

-Dan

Compute the slope of the tangent at the point of contact say $\displaystyle (\alpha,\beta)$which is $\displaystyle \frac{dy}{dx} = \frac{-16\alpha}{9\beta}$. Now we have the slope of the line also as $\displaystyle \frac{6-\beta}{5-\alpha} = \frac{-16\alpha}{9\beta} \implies 16\alpha^2 + 9 \beta^2 = 9*6\beta+9*10\alpha$ from the equation of ellipse we have $\displaystyle 16\alpha^2 + 9 \beta^2 = 9*16=9*6\beta+9*10\alpha \implies 5\alpha+3\beta=8$ substitute in $\displaystyle \frac{\alpha^2}{9} + \frac{\beta^2}{16} = 1 $ and solve for $\displaystyle \beta$ you will get two points on the ellipse, and hence you can get two lines.

~Kalyan.

I had already figured it out but THANKS.