# Equation of tangents

1. Feb 1, 2014

### smart_worker

1. The problem statement, all variables and given/known data

Find the equations of the two tangents that can be drawn from
the point (5, 2) to the ellipse 2x2 + 7y2 = 14

2. Relevant equations

2x2 + 7y2 = 14
differentiating we get slope = -2x/7y

3. The attempt at a solution

y = slope(x)+c
y = -2x2/7y +c

when we plug in x and y we get c=39/7

so 5x+7y=39 is coming as answer

but it seems to be wrong.the questions says there has to be two tangents but how?

x − y = 3 and x − 9y + 13 = 0.

puzzled i dont know how

2. Feb 1, 2014

### SteamKing

Staff Emeritus
Draw a circle. Put a point somewhere outside the circle. How many tangents to the circle can you draw which pass thru the point outside the circle?

3. Feb 1, 2014

### Staff: Mentor

This is the slope at a point on the ellipse, not at the given point (5, 2). If you sketch the ellipse you'll see that there are two points that have tangents that go to (5, 2).
Since the question involves calculus I have moved it to the Calculus & Beyond section.

4. Feb 1, 2014

### smart_worker

ok but now,what should i do to solve this?

5. Feb 1, 2014

### Staff: Mentor

Any point on the ellipse has coordinates (x, $\pm \sqrt{2 - (2/7)x^2}$). Write an expression for the slope between a point on the ellipse and (5, 2), and set it equal to the slope you already found. You should get two x values.

6. Feb 1, 2014

### vela

Staff Emeritus
You have to understand the meaning of the mathematics here. You can't just push symbols around, substituting one quantity for another just because they're both labelled "x" or "y" in different contexts or just because they're both referred to as a slope.

7. Feb 2, 2014

### smart_worker

nope i am getting complex roots

8. Feb 2, 2014

### vela

Staff Emeritus
How? Are you setting x=5 in $\sqrt{2-(2/7)x^2}$?

9. Apr 8, 2014

### smart_worker

hello vela,
i used the slope formula:
(2-$\sqrt{2-(2/7)x^2}$)/(5-x) = -2x/7($\sqrt{2-(2/7)x^2}$)

10. Apr 8, 2014

### HallsofIvy

Staff Emeritus
That does not answer vela's question. Did you set x= 5 in that formula as you were told? Doing that you will NOT get a complex number.

11. Apr 8, 2014

### vela

Staff Emeritus
Actually, if you do that, you will get a complex number. I figured that's how the OP was managing to get complex roots.

This is correct and get you one of the lines. You shouldn't be getting complex roots, however. Without seeing your subsequent work, we can't tell where you're going wrong.

12. Apr 8, 2014

### smart_worker

yes i am getting 7/3 and -7/13.Now can i find y and plug it into the slope.

Last edited: Apr 8, 2014