VCE Specialist 3/4 Question Thread!

[Jaswinder]

smartt! thanks Jenny_2108

[duhherro]

Could anyone help me find the asymtotes with the equation...

4(x-1)^2 /3   - (y-1)^2 /3 = 1

I've tried to use the y-k = plus/minus b(x-h)/a

But not sure how to get by with a 4/3 :/

[Stick]

Hint: Make the denominator of the first expression 3/4.

[duhherro]

Hint: Make the denominator of the first expression 3/4.

Thanks for that! What a nice trick to simply make the denominators in your favour

[b^3]

Hint: The following two expressions are the same

Spoiler

So our asymptotes will be

EDIT: Beaten

[Mr. Study]

Just a trick I found, last year, when finding asymptotes for hyperbolas.

Spoiler

or

or

I find this 'method' easier than remembering y-k = plus/minus b(x-h)/a, as I don't jumble h,a,k,b or w/e.

EDIT: Beaten haha

[shaiga95]

Reduce to following to partial fractions I keep getting it wrong

[brightsky]

long divide first. (in general, if the degree of the numerator is not less than that of the denominator, then long divide before performing partial decomp.)

[b^3]

The working if you really are still stuck, but try to do it on your own first

Spoiler

[shaiga95]

thanks i cross multiplied A and B silly me

[bucklr]

Hi guys,
I came accross this question, I thought I had the answer but I didn't:
For pi/2<A<pi and 0<B<pi/2  sinA=t and cosB=t, cos(B+A)=??

i got 0 but the answer says -2t(1-t^2)^(1/2)
How did they get this?
Many thanks.

[Limista]

Hi guys,
I came accross this question, I thought I had the answer but I didn't:
For pi/2<A<pi and 0<B<pi/2  sinA=t and cosB=t, cos(B+A)=??

i got 0 but the answer says -2t(1-t^2)^(1/2)
How did they get this?
Many thanks.

pi/2<A<pi is in the second quadrant.
0<B<pi/2 is in the first quadrant.

cos (B + A) = cosBcosA - sinBsinA
We need to work out what cosA and sinB is equal to.

cosA = -(1-t²)^(1/2)......why?
sinA = t, so opposite side of triangle = t, hypotenuse of triangle = 1, and adjacent side of triangle = (1/t²)^(1/2). Because A is in the 2nd quadrant and cosine is negative in the 2nd quadrant, cosA = -(1-t²)^(1/2)......

sinB = (1-t²)^(1/2)....why?
follow the same logic as above, except this time sinB will remain positive, because B is in the 1st quadrant and sine is positive in the 1st quadrant.

Substituting these values into the cos (B + A) formula....

t x (-(1-t²)^(1/2)) - (1-t²)^(1/2) x t
= -2t(1-t²)^(1/2)

[bucklr]

Thanks... LOL i thought they were the same triangle. I must stop assuming everything.

Thanks again.

[barydos]

I have some problems with this question.
Could someone help me

[Stick]

OK, so ABD forms an isosceles triangle if you draw a chord between B and D. After that, you'll see angle ADB is also x, and then by utilising the tangent/chord angle in the opposite segment rule, you'll also see angle BCD is x.

For part b, you'll need to use the cross over of two secants rule so that you get y^2=a(a+b).