A conics question (1 Viewer)

[kooltrainer]

find the equation of the tangent to the hyperbola x^2-4y^2 = 16, given that these tangents are parallel to the line x- y root 3 +3 = 0

i dunno how to do this using cartesian equation of tangent .. need help.. thx

 

[Trebla]

x - y√3 + 3 = 0
Gradient of this line is: 1 / √3
Since the tangents are required to be parallel to this line, the gradient of the tangents must be 1 / √3

x² - 4y² = 16 is the equation of the hyperbola
Let the equation of the tangents be generalised by y = mx + c
Since the tangents touch the hyperbola we can sub in y = mx + c into the equation of the hyperbola
x² - 4(mx + c)² = 16
x² - 4(m²x² + 2mxc + c²) = 16
x²(1 - 4m²) - 8mc.x + [- 4c² - 16] = 0
BUT a tangent must touch the hyperbola at ONE point only so there is only ONE solution to the quadratic equation hence Δ = 0
64m²c² - 4(1 - 4m²)(- 4c² - 16) = 0
64m²c² + 16(1 - 4m²)(c² + 4) = 0
4m²c² + (c² + 4 - 4m²c² - 16m²) = 0
4m²c² + c² + 4 - 4m²c² - 16m² = 0
.: 16m² = c² + 4
Now we know that m = 1 / √3, so solve for c
c² = 16 / 3 - 4
c² = 4 / 3
.: c = ± 2 / √3
So the equation of the tangents are y = (x ± 2) / √3

lol I think I did the long method...

 

[leoyh]

firstly, differentiate the equation to get dy/dx --> that should come out as x/4y

afterwards, read off x - y root 3 + 3 = 0 for your gradient. This must be equal to x/4y since x/4y is the gradient function of your hyperbola i.e. x/4y = -4y / root 3

from that, make either x or y the subject, and sub that into your original hyperbola and solve for whatever variable you have. after that, sub those values to find what the other variable is i.e. if you made y the subject, sub into the hyperbola and solve for y, then sub it to find what x is.

after that, you have your 2 points of intersection of your tangents with the hyperbola, which should be (-8, 2 root 3) and (8, -2 root 3).

from that, just simply use y-y1 = m(x-x1) where (x1,y1) are the points that i just mentioned above and m = -1/ root 3 since your tangents are parallel to x- y root 3 +3 = 0

final answer --> y root 3 = -x-2 and y root 3 = -x + 2

 

[kooltrainer]

x - y√3 + 3 = 0
Gradient of this line is: 1 / √3
Since the tangents are required to be parallel to this line, the gradient of the tangents must be 1 / √3

x² - 4y² = 16 is the equation of the hyperbola
Let the equation of the tangents be generalised by y = mx + c
Since the tangents touch the hyperbola we can sub in y = mx + c into the equation of the hyperbola
x² - 4(mx + c)² = 16
x² - 4(m²x² + 2mxc + c²) = 16
x²(1 - 4m²) - 8mc.x + [- 4c² - 16] = 0
BUT a tangent must touch the hyperbola at ONE point only so there is only ONE solution to the quadratic equation hence Δ = 0
64m²c² - 4(1 - 4m²)(- 4c² - 16) = 0
64m²c² + 16(1 - 4m²)(c² + 4) = 0
4m²c² + (c² + 4 - 4m²c² - 16m²) = 0
4m²c² + c² + 4 - 4m²c² - 16m² = 0
.: 16m² = c² + 4
Now we know that m = 1 / √3, so solve for c
c² = 16 / 3 - 4
c² = 4 / 3
.: c = ± 2 / √3
So the equation of the tangents are y = (x ± 2) / √3

lol I think I did the long method...

thx.. thats the way i did it as well.. Theres another faster way, like leoyh did