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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

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i guess we need the derivative first
\[2x+14yy'=0\] \[y'=-\frac{x}{7y}\]
This is a wonderful problem satellite

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Take a look, the point (m,0) may not lie on the ellipse, are you getting the complexity?
oh then i probably messed up somewhere
I didn't mean you messed, I just wanted to let you know that the question is good
ive been trying to get this problem for the last like hour
Let \(y=mx+c\) be the tangent to the ellipse and let the point \((m,0) \) also lie on it
so we have to find the equation of the line through (m,0) that touches the ellipse right?
\[y=cx-m\]since we cannot use m for the slope
Right. we can't use m, since its already up there
and we have to make sure that \[c=-\frac{x}{7y}\] as well so perhaps we will get two equations
OK, so lets assume that the line \(y=kx+c\) is a tangent to the given ellipse
i am lost already
So we know c must be equal to \(\pm \sqrt{a^2k^2+b^2}\)
oh oopos
Where a and b are from the ellipse
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 \[c=-\frac{x}{7y}\]
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
So what I am stating up there, is the condition for the line \(y=kx+c\) to be a tangent to that ellipse
And I am considering the general ellipse, \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
ok i have this, tell me what you think
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 \[xm = 8\]
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 \[c=\pm \sqrt{a^2k^2+b^2}\]
So the equation of the line is \[y=kx+\pm \sqrt{a^2k^2+b^2}\]
Also since the line passes through the point (m,0) \[0=km\pm \sqrt{a^2k^2+b^2}\]
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
ok thanks

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