## sandman1 3 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

1. satellite73

i guess we need the derivative first

2. satellite73

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

3. 2bornot2b

This is a wonderful problem satellite

4. 2bornot2b

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

5. satellite73

oh then i probably messed up somewhere

6. 2bornot2b

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

7. sandman1

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

8. 2bornot2b

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

9. satellite73

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

10. 2bornot2b

Right

11. satellite73

$y=cx-m$since we cannot use m for the slope

12. 2bornot2b

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

13. satellite73

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

14. 2bornot2b

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

15. satellite73

16. 2bornot2b

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

17. satellite73

oh oopos

18. 2bornot2b

Where a and b are from the ellipse

19. satellite73

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 $c=-\frac{x}{7y}$

20. 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

21. 2bornot2b

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

22. 2bornot2b

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

23. satellite73

ok i have this, tell me what you think

24. satellite73

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 $xm = 8$

25. 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 $c=\pm \sqrt{a^2k^2+b^2}$

26. 2bornot2b

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

27. 2bornot2b

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

28. 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

29. sandman1

ok thanks