FP1 Rectangular HyperbolaWatch

#1
Quick Question;

A Rectangular Hyperbola, R, has equation xy=6

the coordinates of the points where the line l with eqaution y=4x+5 intersects R are, (-2,-3) and (3/4,8)

When x=3, the equation for the normal N to R is 2y=3x-5

Now this where i need help

find the co-ordinates of the point D, where l and N meet?

isn't it just this y=4x+5 and 2y=3x-5 made = to each other. I get x as 1. whereas it should be (-3,-7)
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8 years ago
#2
(Original post by emmaxoxo)
Quick Question;

A Rectangular Hyperbola, R, has equation xy=6

the coordinates of the points where the line l with eqaution y=4x+5 intersects R are, (-2,-3) and (3/4,8)

When x=3, the equation for the normal N to R is 2y=3x-5

Now this where i need help

find the co-ordinates of the point D, where l and N meet?

isn't it just this y=4x+5 and 2y=3x-5 made = to each other. I get x as 1. whereas it should be (-3,-7)
Yes, you're right about making them equal to each other to find the intersections BUT 2y =/= y. What do you have to do with one of the equations to make them equal?
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#3
(Original post by Clarity Incognito)
Yes, you're right about making them equal to each other to find the intersections BUT 2y =/= y. What do you have to do with one of the equations to make them equal?
yes, i did that.

2(4x+5)=3x-5
8x+10=3x-5
5x=5
x=1

whereas it should be -3
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8 years ago
#4
(Original post by emmaxoxo)
yes, i did that.

2(4x+5)=3x-5
8x+10=3x-5
5x=5

x=1

whereas it should be -3
8x-3x=-10-5
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8 years ago
#5
(Original post by emmaxoxo)
yes, i did that.

2(4x+5)=3x-5
8x+10=3x-5
5x=5

x=1

whereas it should be -3
Last time I looked -5 - 10 is not 5
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#6
Duh, I'm such an idiot. Thanks guys.
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#7
(Original post by Clarity Incognito)
Last time I looked -5 - 10 is not 5
Ok New question

Parabola C has equation y^2=8x and Focus F

P (2,4) and Q(8,-8) intersect C where the line with eqauation y=8-2x

Given cosPFQ=-4/5

use the focus directrix property of C to show PQ has a lenght of 6root5.

i can do it with just the P and Q co-ordinates, but has do u use the focus directrix property of C, which is a=1/2 isnt it?
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8 years ago
#8
(Original post by emmaxoxo)
Ok New question

Parabola C has equation y^2=8x and Focus F

P (2,4) and Q(8,-8) intersect C where the line with eqauation y=8-2x

Given cosPFQ=-4/5

use the focus directrix property of C to show PQ has a lenght of 6root5.

i can do it with just the P and Q co-ordinates, but has do u use the focus directrix property of C, which is a=1/2 isnt it?
Directrix of C is x = -2 Or x + 2 = 0 y^2 = 8x is similar to y^2 = 4ax where a is 2 so the directrix is x + 2 = 0 and the focus is at (2,0)
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#9
(Original post by Clarity Incognito)
Directrix of C is x = -2 Or x + 2 = 0 y^2 = 8x is similar to y^2 = 4ax where a is 2 so the directrix is x + 2 = 0 and the focus is at (2,0)
so how do you use the x=-2, to show that PQ=6root5
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8 years ago
#10
(Original post by emmaxoxo)
so how do you use the x=-2, to show that PQ=6root5
Think about what you need when using the cosine rule. They've given you the angle opposite the side you want and so you need the length of the other two sides of the triangle. What link is there between the distance of the focus to the points and the points to the directrix?
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8 years ago
#11
(Original post by emmaxoxo)
so how do you use the x=-2, to show that PQ=6root5
I suppose you'd know that PF is 4 units, using the focus-directrix property. Then you can work out what FQ is and finally use the cosine rule to work out PQ.

I think it's a stupid question though...
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