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anonymous
 4 years ago
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
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess we need the derivative first

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2x+14yy'=0\] \[y'=\frac{x}{7y}\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1This is a wonderful problem satellite

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Take a look, the point (m,0) may not lie on the ellipse, are you getting the complexity?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh then i probably messed up somewhere

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1I didn't mean you messed, I just wanted to let you know that the question is good

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ive been trying to get this problem for the last like hour

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Let \(y=mx+c\) be the tangent to the ellipse and let the point \((m,0) \) also lie on it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we have to find the equation of the line through (m,0) that touches the ellipse right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=cxm\]since we cannot use m for the slope

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Right. we can't use m, since its already up there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and we have to make sure that \[c=\frac{x}{7y}\] as well so perhaps we will get two equations

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1OK, so lets assume that the line \(y=kx+c\) is a tangent to the given ellipse

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1So we know c must be equal to \(\pm \sqrt{a^2k^2+b^2}\)

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Where a and b are from the ellipse

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are way ahead of me. i only know that the line must look like \[y=c(xm)\] where c is the slope. i also know that we must have \[c=\frac{x}{7y}\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, 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
 4 years ago
Best ResponseYou've already chosen the best response.1So what I am stating up there, is the condition for the line \(y=kx+c\) to be a tangent to that ellipse

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1And I am considering the general ellipse, \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i have this, tell me what you think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we know the line has slope \[\frac{x}{7y}\] and it passes through (m,0) meaning it has the equation \[y=\frac{x}{7y}(xm)\] now we get \[7y^2=x(xm)\] \[7y^2=x^2+xm\] \[x^2+7y^2=xm\] and we also know that on the ellipse \[x^2+7y^2=8\] making \[xm = 8\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1So 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
 4 years ago
Best ResponseYou've already chosen the best response.1So the equation of the line is \[y=kx+\pm \sqrt{a^2k^2+b^2}\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1Also since the line passes through the point (m,0) \[0=km\pm \sqrt{a^2k^2+b^2}\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.1The 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
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