Conics Help Please (1 Viewer)

[goobi]

Find the equations of the tangent and normal to the ellipse at the point .
If the tangent meets the x-axis at P and the normal meets the y-axis at Q, show that PQ touches the ellipse.

Edit:

So I've found the equation of the tangent:
......(1)

And the equation of the normal:
......(2)

Substituting y=0 into (1), P is

Substituting x=0 into (2), Q is

Gradient of PQ =

Therefore, equation of PQ is:

And I have no idea what to do next...

Any help would be appreciated

 

[nightweaver066]

Find P by letting y = 0 in the tangent equation, P(2a, 0)

Find Q by letting x = 0 in the normal equation,

Find the equation of PQ.

Sub it in the equation of ellipse.

Discriminant = 0 (or factorise it in to a perfect square) therefore touches it.

Unsure if it's the quickest method.

 

[Shadowdude]

^ That.

 

[nightweaver066]

^ That.

lol Shadowdude. Sorry for posting before you.

 

[goobi]

Find P by letting y = 0 in the tangent equation, P(2a, 0)

Find Q by letting x = 0 in the normal equation,

Find the equation of PQ.

Sub it in the equation of ellipse.

Discriminant = 0 (or factorise it in to a perfect square) therefore touches it.

Unsure if it's the quickest method.

Thanks!

 

[nightweaver066]

 

[Aesytic]

the LHS isn't supposed to equal the RHS,
when you sub PQ into the ellipse, you'll get:
x^2 + 3*[(1/root3)*(x-2a)]^2 = 2a^2
x^2 + 3*[1/3*(x-2a)^2] = 2a^2
x^2 + (x-2a)^2 = 2a^2
x^2 + x^2 - 4ax + 4a^2 = 2a^2
2x^2 - 4ax + 2a^2 = 0
2(x^2 - 2ax + a^2) = 0
2(x-a)^2 = 0
since this eqn. has only one solution, there is only one point where PQ and the ellipse meet, .'. the line PQ is a tangent to the ellipse

 

[goobi]

the LHS isn't supposed to equal the RHS,
when you sub PQ into the ellipse, you'll get:
x^2 + 3*[(1/root3)*(x-2a)]^2 = 2a^2
x^2 + 3*[1/3*(x-2a)^2] = 2a^2
x^2 + (x-2a)^2 = 2a^2
x^2 + x^2 - 4ax + 4a^2 = 2a^2
2x^2 - 4ax + 2a^2 = 0
2(x^2 - 2ax + a^2) = 0
2(x-a)^2 = 0
since this eqn. has only one solution, there is only one point where PQ and the ellipse meet, .'. the line PQ is a tangent to the ellipse

Repped.

 

[Shadowdude]

lol Shadowdude. Sorry for posting before you.

Don't worry about it. What I wrote was pretty much the same as yours anyway, but in a more lazy fashion!

 

[math man]

Find the equations of the tangent and normal to the ellipse at the point .
If the tangent meets the x-axis at P and the normal meets the y-axis at Q, show that PQ touches the ellipse.

Edit:

So I've found the equation of the tangent:
......(1)

And the equation of the normal:
......(2)

Substituting y=0 into (1), P is

Substituting x=0 into (2), Q is

Gradient of PQ =

Therefore, equation of PQ is:

And I have no idea what to do next...

Any help would be appreciated

Whenever it says to prove a line touches a curve it means prove it is a tangent and in general the easiest and most straight forward method
is to sub the straight line into the curve and form a quadratic and then show the discriminant equals 0 as this proves the line and curve only
intersect at one point (i.e. it is a tangent). You were lucky for this question that you could factorise easily and show there was only one root.
For future questions of this type showing discriminant = 0 is the way to go.

 

[goobi]

Whenever it says to prove a line touches a curve it means prove it is a tangent and in general the easiest and most straight forward method
is to sub the straight line into the curve and form a quadratic and then show the discriminant equals 0 as this proves the line and curve only
intersect at one point (i.e. it is a tangent). You were lucky for this question that you could factorise easily and show there was only one root.
For future questions of this type showing discriminant = 0 is the way to go.

I saw your post just now but thanks