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Maths questions (1 Viewer)

Smile12345

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Hello All...

Not sure how to do the following:

"X is a point on the curve . Point Y lies directly below X and is on the curve ."

a) Show that the distance d between X and Y is given by .

b) Find the minimum distance between X and Y.

AND

'Points A (-a, 0), B (0,b) and O (0,0) form a triangle and AB always passes through the point (-1,2)'

a) Show that b = .
b) Find values of a and b that give the minimum area of triangle OAB.

Thanks. :D
 

Squar3root

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1)
a) perpendicular distance formula

b) differentate a) and equate it with zero then solve

2)
a) no idea
 

HeroicPandas

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Q1
a) Y is directly below X, so to find distance XY, just get top point minus bottom point (y-values) (XY is a vertical distance)
Before this is all done, tell me what point X and Y are

b) Calculus

Q2
a) U can do this, use what they gave u, the point (-1,2) lies on AB --> what does this mean...?

b) To find 2 unknowns, u need 2 equations. Make another equation with 'a' and 'b' (i.e. a relation between a and b) somehow...
If (-1,2) lies on the line AB, what else can u deduce besides that (-1,2) satisfies eqn AB? (it'll be helpful if you can produce a diagram to inspect)
 
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obliviousninja

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2a) Find equation of AB. Sub in (-2,1), then make b the subject.
 

anomalousdecay

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my bad, the maths has fallen out of my head :p

1)a)
since they have the same x ordinate



b)




lol dont know how to use latex
Fixed it for you. Just put \frac{numerator}{denominator} and put \text{your text here}
These text wrappings make the difference.
 

HeroicPandas

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my bad, the maths has fallen out of my head :p

1)a)
since they have the same x ordinate



b)




lol dont know how to use latex
i dont understand your working out for part (a)...how is d = 'blah' when their x-coordinates are the same? (d is the distance between X and Y)
 

Squar3root

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Fixed it for you. Just put \frac{numerator}{denominator} and put \text{your text here}
These text wrappings make the difference.
Thanks for that. +1 ;)

i dont understand your working out for part (a)...how is d = 'blah' when their x-coordinates are the same? (d is the distance between X and Y)
If the x coordinate is the same (given) then we can equate the 2 functions. I hope this makes sense
 

HeroicPandas

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If the x coordinate is the same (given) then we can equate the 2 functions. I hope this makes sense
nah, i'm asking how did u deduce that d = 'blah' with that equivalency (x^2-2x+5 = 4x-x^2)

Below is invisible because i dont want Smile12345 to see it yet:

so, say the x-coord of point X and Y is x = x1, now the points X and Y become: (taking into account that they have the same x-coord)

Then X(x1, 4x1 - x1^2)

Then Y(x1, x1^2 - 2x1 + 5)

By the looks of it, u equated the y-coordinates!! And if they ALSO have the same y-coord, then X and Y are the same point and so the distance between X and Y is 0

 
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Squar3root

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nah, i'm asking how did u deduce that d = 'blah' with that equivalency (x^2-2x+5 = 4x-x^2)

Below is invisible because i dont want Smile12345 to see it yet:

so, say the x-coord of point X and Y is x = x1, now the points X and Y become: (taking into account that they have the same x-coord)

Then X(x1, 4x1 - x1^2)

Then Y(x1, x1^2 - 2x1 + 5)

By the looks of it, u equated the y-coordinates!! And if they ALSO have the same y-coord, then X and Y are the same point and so the distance between X and Y is 0

Umm, algebra. I just moved everything to one side and that's the distance function
 

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