Yahoo Answers: Answers and Comments for F.5 Maths Module 2 question [數學]
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From Pak Ho
zhHantHK
Thu, 12 Dec 2013 19:58:29 +0000
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Yahoo Answers: Answers and Comments for F.5 Maths Module 2 question [數學]
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From 少年時:
x^2 + 4y^2 = 68
Differentiate both sides w.r....
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https://hk.answers.yahoo.com/question/index?qid=20131212000051KK00163
Sat, 14 Dec 2013 13:55:03 +0000
x^2 + 4y^2 = 68
Differentiate both sides w.r.t. x, we get
2x + 8y dy/dx = 0
==> dy/dx = x/(4y)
So, the slope of the normal at any point on this curve is 4y/x.
Suppose (x, y) is the required point on this curve, then
(y  3)/(x  0) = 4y/x
==> x(y  3) = 4xy
==> x(y  3)  4xy = 0
==> x(y  3  4y) = 0
==> 3x(y + 1) = 0
==> x = 0 or y = 1
When x = 0,
4y^2 = 68
==> y = √17 or y = √17When y = 1,x^2 + 4(1)^2 = 68==> x = 8 or 8equation of normal passes thru' (0, 3) and (0, √17) is x = 0;equation of normal passes thru' (0, 3) and (0, √17) is x = 0;equation of normal passes thru' (0, 3) and (8, 1) is :(y  3)/(x  0) = 4*(1)/8==> x + 2y  6 = 0equation of normal passes thru' (0, 3) and (8, 1) is :(y  3)/(x  0) = 4*(1)/(8)==> x  2y + 6 = 0
Therefore, the equation of normals are :(i) x = 0(ii) x + 2y  6 = 0(iii) x  2y + 6 = 0
(Totally there are 3 normals pass thru' (0, 3))

From Shang Him: x^2+〖4y〗^2=68
2x+4(2y)dy/dx=0
dy/dx=(2x...
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https://hk.answers.yahoo.com/question/index?qid=20131212000051KK00163
Sat, 14 Dec 2013 13:22:08 +0000
x^2+〖4y〗^2=68
2x+4(2y)dy/dx=0
dy/dx=(2x)/8y
dy/dx=(x)/4y
When x=0, y=3
dy/dx=(0)/3=0
Equation of the normal to C:
y3=(1)/0(x0)
y=3

From CK: Refer to : http://www.mathportal.org/formulas/...
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https://hk.answers.yahoo.com/question/index?qid=20131212000051KK00163
Thu, 12 Dec 2013 21:05:00 +0000
Refer to : http://www.mathportal.org/formulas/analyticgeometry/conic.php
to get the equation of tangent.
The ellipse is :
x²/(√68)² + y²/(√17)² = 1
Tangent at (0,3) is 0x/(√68)² + 3y/(√17)² = 1
3y = 17
slope of tangent = 0
slope of normal is ∞
equation of normal at (0,3) is x=0;