Tangent lines to an ellipse from an external point

**Kaloda** · Sep 4th 2012, 03:45 AM

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

**topsquark** · Sep 4th 2012, 05:54 AM

Originally Posted by Kaloda

Given an the equation of an ellipse, say $\frac{x^2}{9}+\frac{y^2}{16}=1$ ,
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

**kalyanram** · Sep 4th 2012, 05:58 AM

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

~Kalyan.

**Kaloda** · Sep 8th 2012, 08:36 AM

I had already figured it out but THANKS.

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