Conics Help (1 Viewer)

[cutemouse]

Hi, I have no idea on how to approach this question. I'd appreciate it if someone could help me. Thank you

3x-2y-1=0 is a chord of the ellipse (what is this?) x²+4y²=9. Find the coordinates of its midpoint.

 

[gurmies]

Wouldn't you just solve the two simultaneously to get two co-ordinates and then use midpoint formula?

 

[Trebla]

Hi, I have no idea on how to approach this question. I'd appreciate it if someone could help me. Thank you

3x-2y-1=0 is a chord of the ellipse (what is this?) x²+4y²=9. Find the coordinates of its midpoint.

Basically you have to find the points where the chord cuts the ellipse and then determine the midpoint.
3x - 2y - 1 = 0 => y = (3x - 1)/2 => 4y² = (3x - 1)²
x² + 4y² = 9
Solve simultaneously:
x² + (3x - 1)² = 9
x² + 9x² - 6x + 1 = 9
5x² - 3x - 4 = 0
We can find the midpoint without having to explicitly find the intersection of the chord and ellipse.
Let the roots of this quadratic be x₁ and x₂. These roots are the x-coordinates of the two intersection points of the chord with the ellipse (x₁, y₁) and (x₂, y₂).
Therefore the midpoint has coordinates M([x₁ + x₂]/2 , [y₁ + y₂]/2)
Sum of roots:
x₁ + x₂ = 3/5
=> (x₁ + x₂)/2 = 3/10
We need (y₁ + y₂)/2, so from equation of chord:
y₁ = (3x₁ - 1)/2
y₂ = (3x₂ - 1)/2
=> y₁ + y₂ = 3(x₁ + x₂)/2 - 1
= 3(3/10) - 1
= -1/10
.: (y₁ + y₂)/2 = - 1/20
.: Midpoint M is (3/10, -1/20)

 

[jet]

A chord of an ellipse is a line joining two of the points on its circumference.

So you would solve the two:
y = (3/2)x - (1/2)

x^2 + 4((3/2)x - (1/2))^2 = 9
x^2 + 4((9x^2)/4 - (3x/2) + (1/4)) = 9
x^2 + 9x^2 - 6x + 1 = 9
10x^2 - 6x - 8 = 0
5x^2 - 3x - 4 = 0

Now, the x-coordinate of the midpoint will be the sum of the roots/2
Hence, x(mp) = ∑α/2 = 3/10

Now, y = (3/2)(3/10) - (1/2)
= 9/20 - 1/2
= -1/20

Hence the midpoint = (3/10, -1/20)

 

[cutemouse]

Hello,

Thanks for your help guys

I have another question. I'm a bit lost on this one. So an explanation of some steps would be great.

a) Find the catesian equation of the ellipse that P(2cosθ, sinθ) and Q(-2sinθ, cosθ) both belong to. (Done this one, it is x²/4 + y²=1)

b) The tangents at P and Q to the ellipse meet each other in R. Show that the locus of R is another ellipse.

Thanks

 

[jet]

Okay, Assuming you know implicit differentiation,
x/2 + 2ydy/dx = 0
2ydy/dx=-x/2
dy/dx=-x/4y

At P, dy/dx =-2cosθ/4sinθ=-cosθ/2sinθ

At Q, dy/dx = 2sinθ/4cosθ = sinθ/2cosθ

Hence, the tangent at P has eqn y - sinθ = -(cosθ/2sinθ)(x-2cosθ)
2ysinθ - 2sin^2(θ) = -xcosθ + 2cos^2(θ)
xcosθ +2ysinθ = 2(sin^2(θ) + cos^2(θ))
xcosθ + 2ysinθ = 2 (1)

The tangent at Q has eqn y - cosθ = (sinθ/2cosθ)(x + 2sinθ)
2ycosθ - 2cos^2(θ) = xsinθ + 2sin^2(θ)
2ycosθ - xsinθ = 2(sin^2(θ) + cos^2(θ))
2ycosθ - xsinθ = 2 (2)

Now, to solve them, we first eliminate the x terms.
(1) x sinθ: 2ysin^2(θ) + xcosθsinθ = 2sinθ (3)

(2) x cosθ: 2ycos^2(θ) - xcosθsinθ = 2cosθ (4)

(3) + (4): 2y=2(sinθ + cosθ)
y = sinθ + cosθ (5)

Doing the same thing to eliminate the y term,
x/2 = (cosθ - sinθ) (6)

Now for the step alot of people dont see:
(5)^2 + (6)^2: (sinθ + cosθ)^2 + (cosθ - sinθ)^2 = (x^2)4 + y^2
(x^2)/4 + y^2 = sin^2(θ) + 2sinθcosθ + cos^2(θ) + cos^2(θ) -2sinθcosθ + sin^2(θ)
(x^2)/4 + y^2 = 2(sin^2(θ) + cos^2(θ))
(x^2)/4 + y^2 = 2
(x^2)/8 + (y^2)/2 = 1

hence the locus of R is an ellipse.

 

[cutemouse]

Now for the step alot of people dont see:
...

Thanks for your help. But I have a question about that... I'm pretty sure you're getting sick of hearing this from me... Why do you square the results that were obtained?

 

[jet]

Just for fun. It helps us lead to the ellipse, and many/most of the trig identities involve squares. With a little bit of insight i could see that the squares would cancel out.