Tangent to an ellipse

**mathieumg** · October 15th 2009, 12:49 AM

I am helping a friend with a calculus problem he has, he just started his first calculus course while I have done a couple in the past. However, it seems like I'm rusty

Here is the problem:

We have an ellipse:

$9\,{x}^{2}+8\,{\it xy}+25\, \left( y-3 \right) ^{2}=36$

And a point not on the ellipse at (2;2).

We want to know the two points on the ellipse where there is a tangent also going through that point.

First, I have implicitly differentiated the equation of the ellipse to end up with:

${y' = \frac {18\,x+8\,y}{-8\,x-50\,y+150}}$

Then, since I know that:

${\frac {2-y}{2-x} = \frac {18\,x+8\,y}{-8\,x-50\,y+150}}$

I found:

${y = -({\frac {36\,x-18\,{x}^{2}+16\,y-8\,yx}{-8\,x-50\,y+150}}-2)}$

Now I need to solve the following system for x and y:

$9\,{x}^{2}+8\,{\it xy}+25\, \left( y-3 \right) ^{2}=36$

${y = -({\frac {36\,x-18\,{x}^{2}+16\,y-8\,yx}{-8\,x-50\,y+150}}-2)}$

But I am unable to. My friend is specifically allowed to use a symbolic calculator to solve the problem, but with Maple I get the following garble:

1 / / 4 3
{ x = - --------------- \1600 RootOf\1600 _Z - 20000 _Z
\ 18 (32 xy - 65)

2 2
+ (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z + 576 xy + 141453

\ / 4 3 2
+ 11024 xy/^3 - 14800 RootOf\1600 _Z - 20000 _Z + (93997 + 512 xy) _Z

2 \ /
+ (-191958 - 5776 xy) _Z + 576 xy + 141453 + 11024 xy/^2 + 45897 RootOf\

4 3 2
1600 _Z - 20000 _Z + (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z

2 \ / 4 3
+ 576 xy + 141453 + 11024 xy/ + 512 RootOf\1600 _Z - 20000 _Z

2 2
+ (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z + 576 xy + 141453

\ \ / 4 3
+ 11024 xy/ xy - 2888 xy - 45279/, y = RootOf\1600 _Z - 20000 _Z

2 2
+ (93997 + 512 xy) _Z + (-191958 - 5776 xy) _Z + 576 xy + 141453

\\
+ 11024 xy/ }

I have tested the two pairs of solution given with both of the equations in my system and they work. I need help to solve that system here and find the two (x;y) pairs. Thank you a lot in advance.

**CaptainBlack** · October 15th 2009, 01:38 AM

Originally Posted by mathieumg

I am helping a friend with a calculus problem he has, he just started his first calculus course while I have done a couple in the past. However, it seems like I'm rusty

Here is the problem:

We have an ellipse:

$9\,{x}^{2}+8\,{\it xy}+25\, \left( y-3 \right) ^{2}=36$

And a point not on the ellipse at (2;2).

The equation of a line through $(2,2)$ with slope $m$ is:

$y=mx-2m+2$

That this is a tangent to the ellipse is equivalent to asking that:

$9x^2+8x(mx-2m+2)+25(mx-2m+2-2)^2=36\ \ \ ... \ (1)$

has only one root, or rather that:

$x^2(25m^2+8m+9)+x(-100m^2-66m+16)+(100m^2+100m-11)=0$

has a single root. This occurs when the discriminant is zero:

$(-100m^2-66m+16)^2-4(25m^2+8m+9)(100m^2+100m-11)=0$

or:

$1136 m^2+1340 m-163=0$

Solve this and you will have the two slopes and hence the two tangent equations.

having forund the slope substitute back into equation $(1)$ to solve for the $x$ coordinate of the point of tangency and substitute that into the equation of the tangent line to find $y$ .

CB

**mathieumg** · October 15th 2009, 01:54 AM

Originally Posted by CaptainBlack

This occurs when the discriminant is zero:

$(-100m^2-66m+16)^-4(25m^2+8m+9))+(100m^2+100m-11)=0$

Shouldn't it be this?

$(-100m^2-66m+16)^2-4(25m^2+8m+9)(100m^2+100m-11)=0$

**CaptainBlack** · October 15th 2009, 02:44 AM

Originally Posted by mathieumg

Shouldn't it be this?

$(-100m^2-66m+16)^2-4(25m^2+8m+9)(100m^2+100m-11)=0$

Yes, finger trouble

CB

**mathieumg** · October 15th 2009, 09:17 AM

Then doesn't it mean we'll end up with $m^4$ and $m^3$ values?

**CaptainBlack** · October 15th 2009, 09:21 AM

Originally Posted by mathieumg

Then doesn't it mean we'll end up with $m^4$ and $m^3$ values?

No they should cancel out

CB

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