- anonymous

Let E be the ellipse given by the equation X^2+7y^2=8
1) if m is any real number, find all tangent lines to E that pass throught the point (m,0)
2) the ellipse E has a tangent line with postive slope that passes throught the point (-8,0) find the point of intersection of this line with the vertical line at x=13 need help please

- katieb

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- anonymous

i guess we need the derivative first

- anonymous

\[2x+14yy'=0\]
\[y'=-\frac{x}{7y}\]

- 2bornot2b

This is a wonderful problem satellite

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## More answers

- 2bornot2b

Take a look, the point (m,0) may not lie on the ellipse, are you getting the complexity?

- anonymous

oh then i probably messed up somewhere

- 2bornot2b

I didn't mean you messed, I just wanted to let you know that the question is good

- anonymous

ive been trying to get this problem for the last like hour

- 2bornot2b

Let \(y=mx+c\) be the tangent to the ellipse and let the point \((m,0) \) also lie on it

- anonymous

so we have to find the equation of the line through (m,0) that touches the ellipse right?

- 2bornot2b

Right

- anonymous

\[y=cx-m\]since we cannot use m for the slope

- 2bornot2b

Right. we can't use m, since its already up there

- anonymous

and we have to make sure that \[c=-\frac{x}{7y}\] as well so perhaps we will get two equations

- 2bornot2b

OK, so lets assume that the line \(y=kx+c\) is a tangent to the given ellipse

- anonymous

i am lost already

- 2bornot2b

So we know c must be equal to \(\pm \sqrt{a^2k^2+b^2}\)

- anonymous

oh oopos

- 2bornot2b

Where a and b are from the ellipse

- anonymous

you are way ahead of me. i only know that the line must look like
\[y=c(x-m)\] where c is the slope.
i also know that we must have
\[c=-\frac{x}{7y}\]

- 2bornot2b

Yes, you are right. And I was thinking of considering that at the end. Because we have two conditions to consider.
First the line is a tangent
And the second is what you are doing, i.e. the point lies on that line

- 2bornot2b

So what I am stating up there, is the condition for the line \(y=kx+c\) to be a tangent to that ellipse

- 2bornot2b

And I am considering the general ellipse, \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

- anonymous

ok i have this, tell me what you think

- anonymous

we know the line has slope
\[-\frac{x}{7y}\] and it passes through (m,0) meaning it has the equation
\[y=-\frac{x}{7y}(x-m)\] now we get
\[7y^2=-x(x-m)\]
\[7y^2=-x^2+xm\]
\[x^2+7y^2=xm\] and we also know that on the ellipse
\[x^2+7y^2=8\] making
\[xm = 8\]

- 2bornot2b

So the if the line y=kx+c is a tangent to \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) then by theorem we know
\[c=\pm \sqrt{a^2k^2+b^2}\]

- 2bornot2b

So the equation of the line is \[y=kx+\pm \sqrt{a^2k^2+b^2}\]

- 2bornot2b

Also since the line passes through the point (m,0)
\[0=km\pm \sqrt{a^2k^2+b^2}\]

- 2bornot2b

The above equation will give you two values of k, by solving the quadratic. Solve them, and put them in the actual equation to the st line, and you will get the two tangents

- anonymous

ok thanks

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