A conics question (1 Viewer)

[let.me.die]

hi. i dont know how to do this question and was wondering if someone could help me out.

1. find the equations of the tangents to the ellipse x^2 + 4y^2 = 9 which are parallel to the line 2x + 3y = 0.

 

[Trebla]

x² + 4y² = 9
It needs to parallel to 2x + 3y = 0 meaning that:
m = - 2/3
So your equation of tangent is y = - (2/3)x + b
But we need to find b so,
Substitute into ellipse
x² + 4(-2x/3 + b)² = 9
x² + 4(4x²/9 - 4bx/3 + b²) = 9
x² + 16x²/9 - 16bx/3 + 4b² = 9
9x² + 16x² - 48bx + 36b² = 81
(25)x² - (48b)x + (36b² - 81) = 0
In order for this to be a tangent it touches the ellipse at one point thus Δ = 0
Δ = 2304b² - 4(25)(36b² - 81)
.: 2304b² - 3600b² + 8100 = 0
1296b² = 8100
b² = 25/4
.: b = ±(5/2)
.: Equations are y = - 2x/3 + 5/2 and y = - 2x/3 - 5/2

 

[Trebla]

Alternative method

Alternatively, this the way most would probably approach it: (in my opinion it's longer but has less algebraic complications)
x² + 4y² = 9
Differentiate implicitly:
2x + 8y.(dy/dx) = 0
dy/dx = - x/4y
It needs to parallel to 2x + 3y = 0 meaning that:
m = - 2/3
.: - 2/3 = - x/4y
2/3 = x/4y
.: y = 3x/8
Substitute into ellipse
x² + 4(9x²/64) = 9
x² + 36x²/64 = 9
x² + 9x²/16 - 9 = 0
16x² + 9x² - 144 = 0
25x² = 144
x² = 144/25
.: x = ±(12/5)
When x = ±(12/5)
y = ±(9/10) respectively
.: The co-ordinates of the point of contacts of the tangents are (12/5 , 9/10) and (- 12/5 , - 9/10)
.: Equation of tangent at (12/5 , 9/10)
y - 9/10 = - 2/3 (x - 12/5)
.: 30y - 27 = 48 - 20x
30y = - 20x + 75
.: y = - 2x/3 + 5/2
.: Equation of tangent at (- 12/5 , - 9/10)
y + 9/10 = - 2/3 (x + 12/5)
30y + 27 = - 20x - 48
30y = - 20x - 75
.: y = - 2x/3 - 5/2

 

[let.me.die]

oooooooooooh. thanks for the help.

i got stuck on another question though.. its from fitzpatrick 32c

13. the chord joining the points Φ and Ө on the ellipse with equation x²/a² +y²/b² = 1 subtend a right angle at the point (a,0). show that:
tan Ө/2 tan Φ/2 = - b² / a²