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Analytical Geometry
27 Oct 2007 23:22:24 IST
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boredom
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Joined: 14 Jan 2007
Posts: 602
27 Oct 2007 23:31:47 IST
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Hry
maximum length of normal would be along the x axis inthis case and hence it would be 3,a+5=3,
a=-2.
Hope it is right........
take care
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28 Oct 2007 00:34:10 IST
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Let's find the normal at any general point (acos
,bsin).
At this point dy/dx = (dy/d) / (dx/d) = bcos/(-asin)
So the slope of normal = -1/(dy/dx) = asin/bcos
Equations of normal : y - bsin = (asin/bcos)(x - acos)
ax.sin - by.cos = (a2-b2)sincos
Centre of ellipse is (0,0) so perpendicular distance from centre to this normal is given by :
d = |(a2-b2)sincos| / 
((asin)2+(-bcos)2
Now divid numerator and denominator by sincos.
d = (a2-b2) /a2sec2+b2cosec2
If we want to maximize d we should minimize its denominator a2sec2+b2cosec2.
Differentiate this and equate it to zero.
2a2sec2tan - 2b2cosec2cot = 0
This gives tan = (b/a)1/2
I have solved this for general ellipse, put a=3 and b=2 as per the question.
tan = (2/3)1/2
Now draw a right angled triangle and find cosec and sec.
cosec = (5/2)1/2 and sec = (5/3)1/2
so, d = (a2-b2) /
a2sec2+b2cosec2
d = (32-22) /32(5/3) + 22(5/2)
d = 5/5 = 1
a+5 = 1
a = -4
,bsin).At this point dy/dx = (dy/d
) / (dx/d) = bcos/(-asin) So the slope of normal = -1/(dy/dx) = asin
/bcosEquations of normal : y - bsin
= (asin/bcos)(x - acos)ax.sin
- by.cos = (a2-b2)sincosCentre of ellipse is (0,0) so perpendicular distance from centre to this normal is given by :
d = |(a2-b2)sin
cos| / ((asin)2+(-bcos)2 Now divid numerator and denominator by sin
cos.d = (a2-b2) /
a2sec2+b2cosec2If we want to maximize d we should minimize its denominator a2sec2
+b2cosec2.Differentiate this and equate it to zero.
2a2sec2
tan - 2b2cosec2cot = 0This gives tan
= (b/a)1/2 I have solved this for general ellipse, put a=3 and b=2 as per the question.
tan
= (2/3)1/2 Now draw a right angled triangle and find cosec
and sec.cosec
= (5/2)1/2 and sec = (5/3)1/2 so, d = (a2-b2) /
a2sec2+b2cosec2d = (32-22) /
32(5/3) + 22(5/2)d = 5/5 = 1
a+5 = 1
a = -4
28 Oct 2007 00:40:48 IST
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But sir, when we solve it first by taking out dy/dx, then using eq. of a straight line,then using length of perpendicular formula.
we get an expression, when we put (3,0) and (0,2) in it, we get a+5 = 4
This is max. length, and minimum length is coming out to be a+5 = 1
Please explain. the expression i am getting is :
Length =[ x12 + (3/2)y1(9-x12)1/2 ] / [ x12 + (9/4)(9-x12) ]1/2
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