question on conics (1 Viewer)

[FD3S-R]

question 6 exercise 3.4 cambridge:

(a) Show that if y=mx + k is a tangent to the rectangular hyperbola xy=c^2, then k^2 + 4mc^2 = 0

(b) Hence find the equations of the tangents from the point (-1,-3) to the rectangular hyperbola xy=4 and find the coordinates of their points of contact.


i can do this question with ellipse and normal hyperbola, but rectangular hyperbola confuses me since i dont understand its parametric form x=ct and y=c/t (why didnt they use x = cp and y = c/p)

thanks

 

[ngai]

question 6 exercise 3.4 cambridge:
i can do this question with ellipse and normal hyperbola, but rectangular hyperbola confuses me since i dont understand its parametric form x=ct and y=c/t (why didnt they use x = cp and y = c/p)

thanks

there is no difference between using ct c/t or cp c/p...
t is the parameter...or p is the parameter...its just a different name

 

[ngai]

question 6 exercise 3.4 cambridge:
(a) Show that if y=mx + k is a tangent to the rectangular hyperbola xy=c^2, then k^2 + 4mc^2 = 0

tangent to hyperbola means when u sub into the hyperbola, there is only one intersection
y=mx+k, xy=c²
so x(mx+k)=c²
which gives mx² + kx - c² = 0
only one solution to this...hence discriminant = 0, so k^2 + 4mc^2 = 0

(b) Hence find the equations of the tangents from the point (-1,-3) to the rectangular hyperbola xy=4 and find the coordinates of their points of contact.

xy=4, so c²=4
if the tangent is y=mx+k, then (-1,-3) lies on this line, so -m+k=-3, ie. m = k+3
k^2 + 4mc^2 = 0
k² + 4(k+3)*4=0
k² + 16k + 48 = 0
k = ...
m = ... etc etc

 

[FD3S-R]

legend thanks mate

actually question wasnt hard just used to doing it with the long trig methods like in ellipse and hyperbola