basic conic Qs (1 Viewer)

[Hikari Clover]

a line y=2x+c cuts a ellipse x^2 + (y^2)/16 = 1
at 2 points , P(x1,y1), Q(x2,y2)

1)show that the length of PQ is |x1-x2| times root 5

2) if PQ is 2root2, find c


thx~~~

 

[ssglain]

i) By distance formula:
PQ² = (x<SUB>1 </SUB>– x<SUB>2</SUB>)² + (y<SUB>1 </SUB>– y<SUB>2</SUB>)²[/COLOR]
= (x<SUB>1 </SUB>– x<SUB>2</SUB>)² + [(2x<SUB>1</SUB> + c)<SUB> </SUB>– (2x<SUB>2</SUB> + c)]²
= (x<SUB>1 </SUB>– x<SUB>2</SUB>)² + (2x<SUB>1 </SUB>– 2x<SUB>2</SUB>)²
= (x<SUB>1 </SUB>– x<SUB>2</SUB>)² + 4(x<SUB>1 </SUB>– x<SUB>2</SUB>)²
= 5(x<SUB>1 </SUB>– x<SUB>2</SUB>)²
.: PQ = √5 |x<SUB>1 </SUB>– x<SUB>2</SUB>|

ii) I can't think of anything but an extremely long-winded method that involves intersecting the two curves in terms of x and using sum of roots to find x<SUB>1 +</SUB> x<SUB>2 </SUB>= -c/5, then doing the same in terms of y to find y<SUB>1 +</SUB> y<SUB>2 </SUB>= 8c/5.
PQ = 2√2 -> x<SUB>1 </SUB>– x<SUB>2 </SUB>= (2√10)/5
We know that PQ has gradient 2 so (x<SUB>1 </SUB>– x<SUB>2</SUB>)/(y<SUB>1 </SUB>– y<SUB>2</SUB>) = 2 -> y<SUB>1 </SUB>– y<SUB>2</SUB> = (4√10)/5

Then of course solve things simultaneous & substitute back into original equation, etc., etc.. There should be an easier way than this. Anyone else?

 

[Hikari Clover]

回复: Re: basic conic Qs

just using distance formula could get the first one,never thought of this way.......

 

[ssglain]

Whoa that font came out big.

I have to admit - the distance formula was not the first thing I used. The natural thing to think about runs along the lines of "Nah, that's 2U. We're talking 4U conics here." So I was halfway through doing that sum of roots thing when I realised that I didn't need to rewrite x<SUB>2</SUB> in terms of x<SUB>1</SUB>. & I still can't think of an easier way to do ii).

 

[Hikari Clover]

回复: Re: basic conic Qs

i tried thousand times with sum of roots things , still not getting the answer

anyway,thx~~~