Coordinate geometry question
[From: ] [author: ] [Date: 11-12-26] [Hit: ]
0) so AP=rt[(a-1)^2+(-3)^2] and AQ=rt[(a-3)^2+(-2)^2].for minimum AP+AQ, differentiate rt[(a-1)^2+(-3)^2]+rt[(a-1)^2+(-3)^2] with respect to a , put the differentiated expression equal to 0 and find the value of a.-A(x,0) Let S= AP+AQ = sqr[(x-1)^2+9]+ sqr[((x-3)^2+ 4]= sqr(x^2-2x+10) + sqr(x^2-6x+13)S = (1/2)(x^2-2x+10)^(-1/2) *(2x-2) + (1/2)(x^2-6x+13)^(-1/2)* (2x-6)=(x-1)/sqr(x^2-2x+10) + (x-3)/sqr(x^2-6x+13)=0(x-1)/sqr(x^2-2x+10) = (3-x)/sqr(x^2-6x+13)(x-1)*sqr(x^2-6x+13)= (3-x)*sqr(x^2-2x+10)(x^2-2x+1)(x^2-6x+13)= (x^2-6x+9)(x^2-2x+10)X^4-8x^3+26x^2-32x+13= x^4-8x^3+31x^2-78x +900= 5x^2-46x +77X=[46+/-sqr(46^2-4*5*77)]/10= (46+/-24)/10x= 7 or 2.......
For obvious reasons, the x value will lie between 1 and 3.
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let the point be (a,0) so AP=rt[(a-1)^2+(-3)^2] and AQ=rt[(a-3)^2+(-2)^2].
for minimum AP+AQ, differentiate rt[(a-1)^2+(-3)^2]+rt[(a-1)^2+(-3)^2] with respect to a , put the
differentiated expression equal to 0 and find the value of a.
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A(x,0)
Let S= AP+AQ = sqr[(x-1)^2+9]+ sqr[((x-3)^2+ 4]
= sqr(x^2-2x+10) + sqr(x^2-6x+13)
S' = (1/2)(x^2-2x+10)^(-1/2) *(2x-2) + (1/2)(x^2-6x+13)^(-1/2)* (2x-6)
=(x-1)/sqr(x^2-2x+10) + (x-3)/sqr(x^2-6x+13)=0
(x-1)/sqr(x^2-2x+10) = (3-x)/sqr(x^2-6x+13)
(x-1)*sqr(x^2-6x+13)= (3-x)*sqr(x^2-2x+10)
(x^2-2x+1)(x^2-6x+13)= (x^2-6x+9)(x^2-2x+10)
X^4-8x^3+26x^2-32x+13= x^4-8x^3+31x^2-78x +90
0= 5x^2-46x +77
X=[46+/-sqr(46^2-4*5*77)]/10
= (46+/-24)/10
x= 7 or 2.2
Now show that (2.2,0) will minimize the distance.
Hoping this helps!
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AP + AQ = sqrt((x-1)^2+9) + sqrt((x-3)^2+4)
Now we need to differentiate sqrt((x-1)^2+9) + sqrt((x-3)^2+4) with respect to x to find the minimum value.
x will be 2.2
Therefore the point is (2.2, 0) .... Answer
