Question

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

Answer 1

i guess we need the derivative first

Answer 2

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

Answer 3

This is a wonderful problem satellite

Answer 4

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

Answer 5

oh then i probably messed up somewhere

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Right

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

i am lost already

Answer 16

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

Answer 17

oh oopos

Answer 18

Where a and b are from the ellipse

Answer 19

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
\[c=-\frac{x}{7y}\]

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

ok i have this, tell me what you think

Answer 24

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

\[xm = 8\]

Answer 25

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
\[c=\pm \sqrt{a^2k^2+b^2}\]

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

ok thanks