Finding two tangents to the ellipse 2x² + 7y² = 14 from point (5,2)

[smart_worker]

Homework Statement

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

Homework Equations

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

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?

can someone please explain this?

x − y = 3 and x − 9y + 13 = 0.

puzzled i don't know how

 

[SteamKing]

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?

 

[Mark44]

Homework Statement

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

Homework Equations

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

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).

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?

can someone please explain this?

x − y = 3 and x − 9y + 13 = 0.

puzzled i don't know how

Since the question involves calculus I have moved it to the Calculus & Beyond section.

 

[smart_worker]

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).

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

 

[Mark44]

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.

 

[vela]

Homework Statement

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

Homework Equations

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

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?

can someone please explain this?

x − y = 3 and x − 9y + 13 = 0.

puzzled i don't know how

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.

 

[smart_worker]

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.

nope i am getting complex roots

 

[vela]

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

 

[smart_worker]

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

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

 

[HallsofIvy]

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.

 

[vela]

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.

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

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

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.

 

[smart_worker]

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