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i guess we need the derivative first

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

This is a wonderful problem satellite

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?

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

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

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)
\[0=km\pm \sqrt{a^2k^2+b^2}\]

ok thanks