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A Few Problems I Need Help With...[AS/A2]

Okay I'm stuck on a few questions:

1. A curve has equation 7x^2 + 48xy - 7y^2 +75=0

A and B are two distinct points on the curve. At each of these points the gradient of the curve is equal to 2/11.

(a) Use implicit differentiation to show that x +2y =0 at the points A and B.

(b) Find the co-ordinates of the points A and B.

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2. Graph of y=x.(sinx)^1/2. It now says that the maximum point on the curve is A.

Show that the x-co ordinate of the point A satisfied the equation:
2tan x +x =0

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3. A circle has equation

x^2 + y^2 - 6x + 8y - 75=0

(a) Write down the co-ordinates of the centre of C, and calculate the radius of C.

(I did this part of the question, but I got confused as to whether to use (y-4)^2 or (y+4)^2)

It then continues with -

A second circle has centre at the point (15,12) and radius 10.
(b) Find the co-ordinates of the point where both of the circles touch

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If anyone can help I'd really appreciate it - thanks in advance... :biggrin:

Reply 1

insparato

3. A circle has equation

x^2 + y^2 - 6x + 8y - 75=0

(a) Write down the co-ordinates of the centre of C, and calculate the radius of C.

(I did this part of the question, but I got confused as to whether to use (y-4)^2 or (y+4)^2)

It then continues with -

A second circle has centre at the point (15,12) and radius 10.
(b) Find the co-ordinates of the point where both of the circles touch

x^2 + y^2 - 6x + 8y - 75=0 ---- completing the square

(x - 3)²- 9 + (y + 4)² - 16 - 75 = 0

(x - 3)² + (y + 4)² = 100

Equation of circle is (x - a)² + (y - b)² = r²

where (a,b) is centre of circle

So centre of circle is (3,-4) and radius = root 100 = 10

b) Both equations have the radius 10 so both touch at one point ie the midpoint between each radius therefore

(x₂ + x₁/2, y₂ + y₁)

therefore (18/2, 8/2) = (9,4) is where they intersect.

Sorry i cant help you i dont know what implicit differentiation is, i cant differentiated trig yet either its like C3+ stuff ive only just done C2.

Reply 2

Libertine

Ash

1. A curve has equation 7x^2 + 48xy - 7y^2 +75=0

A and B are two distinct points on the curve. At each of these points the gradient of the curve is equal to 2/11.

(a) Use implicit differentiation to show that x +2y =0 at the points A and B.

(b) Find the co-ordinates of the points A and B.

7x² + 48xy - 7y² + 75 = 0
14x + 48y + 48x(dy/dx) - 14y (dy/dx) = 0
7x + 24y + 24x(dy/dx) - 7y (dy/dx) = 0
(dy/dx)(7y-24x) = 7x + 24y
dy/dx = (7x + 24y)/(7y - 24x)
It equals 2/11 at point A and B
(7x + 24y)/(7y - 24x) = 2/11
11(7x + 24y)=2(7y - 24x)
77x + 264y = 14y - 48x
125 + 250y = 9
x + 2y = 0
(there's probably an easier way of doing that)
7x² + 48xy - 7y² + 75 = 0
x = -2y and y=-(x/2)
Substitute them in and solve to get x and y in the different cases.

Reply 3

gyrase

Ash

7x^2 + 48xy - 7y^2 + 75 = 0
=> 14x + 48x(dy/dx) + 48y - 14y(dy/dx) = 0

(a) sub dy/dx = 2/11

=> 14x + 48x(2/11) + 48y - 14y(2/11) = 0
=> 14x + 96x/11 + 48y - 28y/11 = 0
=> 7x + 48x/11 + 24y - 14y/11 = 0
=> 125x/11 + 250y/11 = 0
=> x + 2y = 0

(b) sub x = -2y into original equation;

7(-2y)^2 + 48(-2y)y - 7y^2 +75=0
=> -75y^2 + 75=0
=> y^2 - 1 =0
=> (y+1)(y-1) =0

so y = 1 and y = -1, sub these into [x] to get x-ordinates;

x + 2(-1) = 0, x = 2
and x + 2(1) = 0, x = -2

:. points with grad 2/11 are (1, -2) and (-1, 2)

Reply 4

Ash

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Thanks guys ...

Reply 5

gyrase

Ash

2. Graph of y=x.(sinx)^1/2. It now says that the maximum point on the curve is A.

Show that the x-co ordinate of the point A satisfied the equation:
2tan x +x = 0

y = x(sinx)^1/2
dy/dx = d[x]/dx.(sinx)^1/2 + d[(sinx)^1/2]/dx.x
........= (sinx)^1/2 + x/2(sinx)^(-1/2)cosx

sub dy/dx = 0

0 = (sinx)^1/2 + x/2(sinx)^(-1/2)cosx
=> 0 = [(sinx) + (x/2)cosx]/(sinx)^1/2 [multiplying by (sinx)^1/2/(sinx)^1/2]
=> 0 = (sinx) + (x/2)cosx
=> 0 = tanx + (x/2) [dividing]
=> 0 = 2tanx + x