anonymous
  • 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
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
i guess we need the derivative first
anonymous
  • anonymous
\[2x+14yy'=0\] \[y'=-\frac{x}{7y}\]
2bornot2b
  • 2bornot2b
This is a wonderful problem satellite

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

2bornot2b
  • 2bornot2b
Take a look, the point (m,0) may not lie on the ellipse, are you getting the complexity?
anonymous
  • anonymous
oh then i probably messed up somewhere
2bornot2b
  • 2bornot2b
I didn't mean you messed, I just wanted to let you know that the question is good
anonymous
  • anonymous
ive been trying to get this problem for the last like hour
2bornot2b
  • 2bornot2b
Let \(y=mx+c\) be the tangent to the ellipse and let the point \((m,0) \) also lie on it
anonymous
  • anonymous
so we have to find the equation of the line through (m,0) that touches the ellipse right?
2bornot2b
  • 2bornot2b
Right
anonymous
  • anonymous
\[y=cx-m\]since we cannot use m for the slope
2bornot2b
  • 2bornot2b
Right. we can't use m, since its already up there
anonymous
  • anonymous
and we have to make sure that \[c=-\frac{x}{7y}\] as well so perhaps we will get two equations
2bornot2b
  • 2bornot2b
OK, so lets assume that the line \(y=kx+c\) is a tangent to the given ellipse
anonymous
  • anonymous
i am lost already
2bornot2b
  • 2bornot2b
So we know c must be equal to \(\pm \sqrt{a^2k^2+b^2}\)
anonymous
  • anonymous
oh oopos
2bornot2b
  • 2bornot2b
Where a and b are from the ellipse
anonymous
  • 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
  • 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
  • 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
  • 2bornot2b
And I am considering the general ellipse, \(\huge \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
anonymous
  • anonymous
ok i have this, tell me what you think
anonymous
  • 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
  • 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
  • 2bornot2b
So the equation of the line is \[y=kx+\pm \sqrt{a^2k^2+b^2}\]
2bornot2b
  • 2bornot2b
Also since the line passes through the point (m,0) \[0=km\pm \sqrt{a^2k^2+b^2}\]
2bornot2b
  • 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
  • anonymous
ok thanks

Looking for something else?

Not the answer you are looking for? Search for more explanations.