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Answers as follows:


1) ( Method 1 )

Slope of tangent = dy/dx

At ( 9a, -6a ), slope of tangent

= ( 4a / 2y ) | ( 9a, -6a )

= 4a / 2( -6a )

= 4a / -12a

= 1/3

Hence, the equation of tangent at ( 9a, -6a ):

y - ( -6a ) = ( -1/3 )[ x - ( 9a ) ]

3( y + 6a ) = -( x - 9a )

3y + 18a = -x + 9a

x + 3y + 9a = 0
============

P.S.: Your ans. must be incorrect since you do not know what is 'a' here.


( Method 2 )

Use the formula with ( x1, y1 ) be the point on the parabola and the equation of tangent being yy1 = 2a ( x + x1 )

Put ( x1, y1 ) = ( 9a, -6a ), we have

y( -6a ) = 2a ( x + 9a )

-3y = x + 9a

x + 3y + 9a = 0
============


2) Noted the condition for a straight line y = mx + c to be a tangent to the standard ellipse is c^2 = b^2 + ( am )^2

Now we have a = 3, b = 4 and m = 4/3 ( slope of given line ).

Hence c^2 = 16 + 16 = 32

c = +/- Sq Root ( 32 ) = +/- 4 Sq Root ( 2 )

Hence, the equation of tangent is:

y = ( 4/3 ) x +/- 4 Sq Root ( 2 )

3y = 4x +/- 12 Sq Root ( 2 )

4x - 3y = +/- 12 Sq Root (2)
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Hope I can help you.



2011-03-06 20:41:26 補充：
1) I simply used a theorem of ellipse only. You can treat it as a fact.

2) This is Slope-Intercept Form of equation format ( y = mx + c ).

For question 1- 100understood! (OKAY!!!)
For question 2- ??????
1) why c^2=b^2+(am)^2, isn't that for ellipse, c^2=a^2-b^2???
2) why the eq of the tan to the ellipse is y=slopex +/- c ???

**Not needed within this week, i can extend time....for yr explanations