LOCUS AND PARABOLA help (1 Viewer)

[jackerino]

Find equation of parabola with vertex (3,-1) and 4x+y-7=0 is a tangent.

Fuck this is annoying me.

 

[jackerino]

OH AND HELP ME WITH THIS ALSO plz,
eliminate the parameter and hence find the cartesian equation of the curve
x=1+2tantheta, y=3sectheta-4

 

[jackerino]

Pleaase help meee omfgg!!!!!!!

 

[Timothy6340]

do u have any of the solutions ??

 

[jackerino]

do u have any of the solutions ??

yea first one is (x-3)^2=y+1, second one is 4(y+4)^2-9(x-1)^2=36

 

[deswa1]

OH AND HELP ME WITH THIS ALSO plz,
eliminate the parameter and hence find the cartesian equation of the curve
x=1+2tantheta, y=3sectheta-4

Note that sec^2(theta)=tan^2(theta) + 1. Try using this to help eliminate theta- don't have time for a full latex sorry. Good luck

 

[jackerino]

Note that sec^2(theta)=tan^2(theta) + 1. Try using this to help eliminate theta- don't have time for a full latex sorry. Good luck

Yea I tried and ended up with 9x*2+27=4y^2+16. Idk how to get into the answers form.

 

[Timothy6340]

OH AND HELP ME WITH THIS ALSO plz,
eliminate the parameter and hence find the cartesian equation of the curve
x=1+2tantheta, y=3sectheta-4

this might be a bit hard to understan but ill give a shot
the only thing relatin sectheta and tantheta is the identity 1+(tantheta)^2=(sectheta)^2

x-1=2tantheta
(x-1)^2=4(tantheta)^2
(x-1)^2/4=(tantheta)^2
((x-1)^2/4)+1=(tantheta)^2+1 ....(1)

y=3sectheta-4
y+4=3sectheta
(y+4)^2=9(sectheta)^2
((y+4)^2/9)= (sectheta)^2 ....(2)

sub (1) into (2)
((y+4)^2/9)=(X-1)^2/4) +1
((y+4)^2/9)-(x-1)^2/4)=1
4(y+4)^2-9(x-1)^2=36

 

[jackerino]

this might be a bit hard to understan but ill give a shot
the only thing relatin sectheta and tantheta is the identity 1+(tantheta)^2=(sectheta)^2

x-1=2tantheta
(x-1)^2=4(tantheta)^2
(x-1)^2/4=(tantheta)^2
((x-1)^2/4)+1=(tantheta)^2+1 ....(1)

y=3sectheta-4
y+4=3sectheta
(y+4)^2=9(sectheta)^2
((y+4)^2/9)= (sectheta)^2 ....(2)

sub (1) into (2)
((y+4)^2/9)=(X-1)^2/4) +1
((y+4)^2/9)-(x-1)^2/4)=1
4(y+4)^2-9(x-1)^2=36

No its perfectly easy to understand dw, thanks man! I see where I went wrong now.

 

[jackerino]

K now a new one....
Find the equation of the chord of the parabola joining the points with parameters:
a) 1 and -3 on x=2t, y=t^2

 

[jackerino]

I figured out its all good.

 

[jackerino]

NOONE HAS ANSWERED THE FIRST QUESTION ON THE FIRST POST! plz help guys! How to do the question on my first post here??

 

[jackerino]

Ive done this ten times, yet I get the wrong answer apparently. I can SWEAR im right here. The answer says its a(p^2+1)

i) Find an expression using the distance formula for PS on the parabola x=2at, y=at^2, and at P its with parameter p. S is the focus, (0,a)

So basically its sqrt ((ap^2-a)^2+(2ap)^2)) right?? and they get a final a(p^2+1), I get something different.

 

[Timske]

Ive done this ten times, yet I get the wrong answer apparently. I can SWEAR im right here. The answer says its a(p^2+1)

i) Find an expression using the distance formula for PS on the parabola x=2at, y=at^2, and at P its with parameter p. S is the focus, (0,a)

So basically its sqrt ((ap^2-a)^2+(2ap)^2)) right?? and they get a final a(p^2+1), I get something different.

Answer is correct, you have done something wrong

 

[deswa1]

Ive done this ten times, yet I get the wrong answer apparently. I can SWEAR im right here. The answer says its a(p^2+1)

i) Find an expression using the distance formula for PS on the parabola x=2at, y=at^2, and at P its with parameter p. S is the focus, (0,a)

So basically its sqrt ((ap^2-a)^2+(2ap)^2)) right?? and they get a final a(p^2+1), I get something different.

What do you get? I just did it very quickly and their answer is correct. The easiest way to do these is to ignore the square root and then root it later- factorise out the a^2, then expand and you should end up with a perfect square, factorise and then root the whole thing to get the answer.

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=PS = \sqrt{(2ap)^2 @plus; (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 @plus; a^2p^4 - 2a^2p^2 @plus; a^2} \\\ =\sqrt{2a^2p^2 @plus; a^2p^4 @plus; a^2} \\\ = \sqrt{a^2(2p^2 @plus; p^4 @plus; 1)} \\\ =\sqrt{a^2(p^2@plus;1)^2} \\\ =a(p^2 @plus; 1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?PS = \sqrt{(2ap)^2 + (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 + a^2p^4 - 2a^2p^2 + a^2} \\\ =\sqrt{2a^2p^2 + a^2p^4 + a^2} \\\ = \sqrt{a^2(2p^2 + p^4 + 1)} \\\ =\sqrt{a^2(p^2+1)^2} \\\ =a(p^2 + 1)" title="PS = \sqrt{(2ap)^2 + (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 + a^2p^4 - 2a^2p^2 + a^2} \\\ =\sqrt{2a^2p^2 + a^2p^4 + a^2} \\\ = \sqrt{a^2(2p^2 + p^4 + 1)} \\\ =\sqrt{a^2(p^2+1)^2} \\\ =a(p^2 + 1)" /></a>

 

[jackerino]

<a href="http://www.codecogs.com/eqnedit.php?latex=PS = \sqrt{(2ap)^2 @plus; (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 @plus; a^2p^4 - 2a^2p^2 @plus; a^2} \\\ =\sqrt{2a^2p^2 @plus; a^2p^4 @plus; a^2} \\\ = \sqrt{a^2(2p^2 @plus; p^4 @plus; 1)} \\\ =\sqrt{a^2(p^2@plus;1)^2} \\\ =a(p^2 @plus; 1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?PS = \sqrt{(2ap)^2 + (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 + a^2p^4 - 2a^2p^2 + a^2} \\\ =\sqrt{2a^2p^2 + a^2p^4 + a^2} \\\ = \sqrt{a^2(2p^2 + p^4 + 1)} \\\ =\sqrt{a^2(p^2+1)^2} \\\ =a(p^2 + 1)" title="PS = \sqrt{(2ap)^2 + (ap^2 - a)^2} \\\ = \sqrt{4a^2p^2 + a^2p^4 - 2a^2p^2 + a^2} \\\ =\sqrt{2a^2p^2 + a^2p^4 + a^2} \\\ = \sqrt{a^2(2p^2 + p^4 + 1)} \\\ =\sqrt{a^2(p^2+1)^2} \\\ =a(p^2 + 1)" /></a>

Ah thanks, I see what I did wrong now, I did sqrt before adding 4a^2p^2 to -2a^2p^2