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amo1
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hi i cant do these 2 questions any 1 who can i would be v v greatful,
thanks in advance
amo1
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Fermat
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Have a look at this thread.

http://www.thestudentroom.co.uk/t88005.html
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evariste
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(Original post by amo1)
hi i cant do these 2 questions any 1 who can i would be v v greatful,
thanks in advance
amo1
4y-3x=20
gives y=5+3x/4 (1)
this meets circle when x^2+25+30x/4+9x^2/16=16
16x^2+400+120x+9x^2=256
25x^2+120x+144=0
(5x+12)^2=0
only 1 point of intersection with circle so tangent.
(1) meets y^2=15x when
25+30x/4+9x^2/16=15x
400+120x+9x^2=240x
9x^2-120x+400=0
(3x-20)^2=0
only 1 point of intersection at x=20/3 y=10
y^2=15x 2ydy/dx=15
so at x=20/3 y=10
dy/dx=15/20
ie, gradient of curve is 3/4 =grad of line

let line be given by y=mx+c (2)
this meets y^2=4ax when
m^2x^2+2mcx+c^2-4ax=0
m^2x^2+x(2mc-4a)+c^2=0
only intersects in one point when
(2mc-4a)^2-4m^2c^2=0
4m^2c^2-16amc+16a^2-4m^2c^2=0
ie (a-mc)=0 (3)
(2) meets x^2=4ay when
x^2=4amx+4ac
x^2-4amx-4ac=0
this has only 1 solution when
16a^2m^2+16ac=0
am^2+c=0 (4)
(3) gives mc=a
so c=a/m
putting into (4) gives am^2+a/m=0
m^3=-1
m=-1
and c=-a.
ie y=-x-a is required line
CHECK
y=-x-a
intersects x^2=4ay when
x^2+4ax+4a^(2)=(x+2a) so x=-2a is x-cord of only point of intersect.
from x^2=4ay
dy/dx=x/2a so at x=-2a
dy/dx=-1
so grad of line=grad of curve.
y=-x-a meets y^2=4ax when
x^2+2ax+a^2-4ax=0
x^2-2ax+a^2=0
(x-a)^2=0 so x=a x-cord of only point of intersect and y=-2a
finally 2ydy/dx=4a so at y=-2a
dy/dx=-1
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Nima
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19.) L: 4y - 3x = 20
C: x^2 + y^2 = 16
P: y^2 = 15x

L: 4y - 3x = 20 ---> y = (20 + 3x)/4
Sub into C:
---> x^2 + [(20 + 3x)(20 + 3x)]/16 = 16
---> 16x^2 + 9x^2 + 120x + 400 - 256 = 0
---> 25x^2 + 120x + 144 = 0
---> Discriminant = 120^2 - 4(25)(144) = 14400 - 14400 = 0
* As Discriminant of above quadratic = 0, L intersects C at 1 point only and hence L is a tangent to C.

Sub into P:
[(20 + 3x)(20 + 3x)]/16 = 15x
---> 9x^2 + 120x + 400 = 240x
---> 9x^2 - 120x + 400 = 0
---> Discriminant = (-120)^2 - 4(9)(400) = 14400 - 14400 = 0
* As Discriminant of above quadratic = 0, L intersects P at 1 point only and hence L is a tangent to P.

20.) As the tangent is a straight line, let it be of the form y = mx + c, where m and c are constants to be determined.

When tangent meets y^2 = 4ax:
---> (mx + c)(mx + c) = 4ax
---> m^2x^2 + 2mcx + c^2 - 4ax = 0
---> m^2x^2 + (2mc - 4a)x + c^2 = 0
* Discriminant = 0
---> (2mc - 4a)(2mc - 4a) - 4(m^2)(c^2) = 0
---> 4m^2c^2 - 16amc + 16a^2 - 4m^2c^2 = 0
---> 16a^2 - 16amc = 0
---> a - mc = 0
---> a = mc

When tangent meets x^2 = 4ay ---> y = (x^2)/4a:
---> (mx + c) = (x^2)/4a
---> 4amx + 4ac = x^2
---> x^2 - 4amx - 4ac = 0
* Discriminant = 0
---> (-4am)^2 - 4(-4ac) = 0
---> 16a^2m^2 + 16ac = 0
---> am^2 + c = 0

Hence: a = mc and am^2 + c = 0
---> c = a/m

Therefore: am^2 + a/m = 0
---> m^2 + 1/m = 0
---> m^3 + 1 = 0
---> m^3 = -1
---> m = -1

Hence: a = -c ---> c = -a

---> Tangent: y = -(x + a)
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amo1
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#5
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thanks nima
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