Graphs Question for my half yearly (1 Viewer)

[kcqn93]

from my half yearly**

Hey guys, I just did my half yearly and there was a really strange question:

The equation of the curve is x^2y^2 - x^2 + y^2 = 0.
a. Show that the numerical value of y is never greater than the corresponding value of x. (3 marks)
b. Show that the numerical value of y is always less than unity. (3 marks)
c. Find the equation of the asymptotes. (2 marks)
d. Find the equations of the tangents at the origin. (2 marks)
e. Sketch the curve. (2 marks)

Thanks in advance.

 

[bachviete]

from my half yearly**

Hey guys, I just did my half yearly and there was a really strange question:

The equation of the curve is x^2y^2 - x^2 + y^2 = 0.
a. Show that the numerical value of y is never greater than the corresponding value of x. (3 marks)
b. Show that the numerical value of y is always less than unity. (3 marks)
c. Find the equation of the asymptotes. (2 marks)
d. Find the equations of the tangents at the origin. (2 marks)
e. Sketch the curve. (2 marks)

Thanks in advance.

(a)



We must show



ie









which is true

(b) Note that because y is plus and minus then the graph is the graph of and the reflection of that over the x axis

Hence if we prove that then we have proved the statement.

Note that because it is a square root



Assume that , then



ie (false)

Therefore



(c)







is an asymptote

By reflection over the x - axis,

y = -1 is also an asymptote

(d) When x = 0, y = 0. Therefore, through the definition of the square root function, the equations of the asymptotes is x = 0

(e) Just use a computer app, its just looks like a large parabola sideways and reflected over the y-axis between -1 < y < 1 . Note not a very accurate description

 

[kcqn93]

ahhhhh thanks!

i couldn't get it in the exam

 

[StuartLee]

Hi mate,

My answer would be:

The equation of the curve is x^2y^2 - x^2 + y^2 = 0.
a. Show that the numerical value of y is never greater than the corresponding value of x. (3 marks)
Ans:
y^2 = x^2 - x^2.y^2
so, y^2 <= x^2 (coz x^2.y^2 always >=0 )
so, root y^2 <= root x^2
so, |y|<=|x|
so, the numerical value of y is never greater than the correspond value of x.

b. Show that the numerical value of y is always less than unity. (3 marks)
Ans:
Make y^2 the subject, y^2 = x^2 / (1 + x^2)
so, y^2 = 1/ (1 + x^-2) <= 1 / 1 (coz x^-2 always >= 0)


c. Find the equation of the asymptotes. (2 marks)
Ans:
since y^2 = 1/ (1 + x^-2)
as x -> infinity, y^2 -> 1
ie. y -> +1 or -1
So, asymptotes are y = 1 and y = -1


d. Find the equations of the tangents at the origin. (2 marks)
Ans:
I think this is not too hard
Consider y = x / root (x^2 + 1)
y' = 1 / (x^2 + 1)^1.5
when x = 0, y'=1
so, the tangent is y - 0 = 1.(x - 0) ie. y = x
Together with y = -x / root (x^2 + 1)
ie. the tangents are:
y = x and y = -x

e. Sketch the curve. (2 marks)
ans: like a sand clock, put horizontally.

Stuart