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[kooltrainer]

1) find the equation of the tangent to the curve xy(x+y)+16=0 at the point on the curve where the gradient is -1

2)the chord AB of a circle of radius r subtends an angle of 2a radians at the centre O. The perimeter of the minor segment AB is K times the perimeter of the triangle OAB. Show that K +(K-1)sin a = a. Use a graphical method to obtain an estimate of "a" in the case when k=1/2

3)A chord AB of a circle makes an angle a at the centre of circle. If the perimeter of the minor segment is one-half the circumference of the circle, show that
2sin (a/2) = (pi/2) - 2a

 

[Trebla]

1) The equation of the curve is x²y + xy² + 16 = 0. Implicitly differentiate wrt x and you get:
2xy + x²(dy/dx) + y² + 2xy(dy/dx) = 0
dy/dx = - (y² + 2xy)/(x² + 2xy)
You want the gradient as – 1 so
(y² + 2xy)/(x² + 2xy) = 1
y² + 2xy = x² + 2xy
x² - y² = 0
(x – y)(x + y) = 0
So: x = y or x = - y
Sub x = y,
2y³ = - 16
y³ = -8
y = -2
corresponding x-value is -2
Sub x = - y
16 = 0
Which is false so x =/= - y (in fact y = - x is an asymptote)
So the point is (- 2, - 2)
So equation of tangent is: y + 2 = - (x + 2)

 

[kooltrainer]

ok

 

[Mark576]

Q2. l_{MINOR ARC AB} = r@ = 2ar

[l_{CHORD AB}]² = r² + r² - 2r².cos(2a) = 2r²(1-cos(2a))

l_{CHORD AB} = 2r.sin(a) [using cos(2a) = 1 - 2sin²(a)]

Hence: K.[2r + 2r.sin(a)] = 2r.sin(a) + 2ar

=> K.[1 + sin(a)] = sin(a) + a

K + K.sin(a) = sin(a) + a

K + K.sin(a) - sin(a) = a

K + (K - 1).sin(a) = a

 

[kooltrainer]

oh, didnt know its perimeter of segment .. daym