Cambridge 4 unit: Conics 3.3 Question 9 (a) (1 Viewer)

[foram]

Cambridge 4 unit: Conics 3.3 Question 9 (a)

I'm up to:
m= [b.sec@]/[a.tan@] ...(1)
b.tan@ - ma.sec@ = k ...(2)

but the worked solution says that the next step is:
(2)<SUP>2</SUP>-(1)<SUP>2</SUP> -> m<SUP>2</SUP>a<SUP>2</SUP>(sec<SUP>2</SUP>@-tan<SUP>2</SUP>@) + b<SUP>2</SUP>(tan<SUP>2</SUP>@-sec<SUP>2</SUP>@) = k<SUP>2</SUP> [I don't understand this step. Does ma=b? How?]

This is confusing me. Can somebody explain this to me?

 

[Mark576]

ma.tan@ = b.sec@ -> ma.tan@ - bsec@ = 0 ...(1)
b.tan@ - ma.sec@ = k ...(2)

(2)² - (1)² -> b².tan²@ - 2mba.tan@sec@ + m²a²sec²@ - m²a².tan²@ + 2mba.sec@tan@ - b².sec²@ = k²

Result follows.

 

[conics2008]

thats trying to prove that a tangent exsist. i prefer the alegebra crap.. its alot longer but you will understand it alot more.. screw trig.

sorry i cant be stuffed posting my working out.

 

[foram]

thats trying to prove that a tangent exsist. i prefer the alegebra crap.. its alot longer but you will understand it alot more.. screw trig.

sorry i cant be stuffed posting my working out.

can you post it up later if you have time please? It will be very helpful for me. conics is still a bit fuzzy for me. ty

 

[foram]

ma.tan@ = b.sec@ -> ma.tan@ - bsec@ = 0 ...(1)
b.tan@ - ma.sec@ = k ...(2)

(2)² - (1)² -> b².tan²@ - 2mba.tan@sec@ + m²a²sec²@ - m²a².tan²@ + 2mba.sec@tan@ - b².sec²@ = k²

Result follows.

thankyou. I didn't think i was supposed to expand it. silly me.

 

[conics2008]

ok buddy can you post up the question. let me refresh my memeory abit.

 

[foram]

9. a) Show that if y = mx + k is a tangent to the hyperbola x<SUP>2</SUP>/a<SUP>2 </SUP>- y<SUP>2</SUP>/b<SUP>2 </SUP>= 1
that m<SUP>2</SUP>a<SUP>2 </SUP>- b<SUP>2 </SUP>= k<SUP>2</SUP>

 

[conics2008]

Hey i will tell you how to start it off.

start by subin in y=mx+k into the equation x^2/a^2-y^2/b^2=1

when your done expand and make it into general form.. that is qudratic form..

ax^2+bx+c=0

then find the discriminate. you know why because the tangent will will exsist is the discriminate is greater then 0... when you do that you will find that m^2a^2-b^2-k^2=0 hence you found what you need.

its going to be abit long working out soo i would suggest you to take your time and be patient.

i cant post up the writing here, i did it on my paper and its fine =] good luck

 

[ronnknee]

then find the discriminate. you know why because the tangent will will exsist is the discriminate is greater then 0... when you do that you will find that m^2a^2-b^2-k^2=0 hence you found what you need.

Wrong

The tangent exists if and only if the discriminant is equal to 0

 

[conics2008]

Wrong

The tangent exists if and only if the discriminant is equal to 0

its same bull shit... greater and equal to 0

 

[ronnknee]

Just equal to 0

 

[ronnknee]

*hi5* 3unitz

 

[tacogym27101990]

its same bull shit... greater and equal to 0

yeah definitely only equal to zero

 

[tommykins]

only = 0

 

[conics2008]

fckk relax i got the point..

LoL....

 

[Paj20]

Def only = to 0!