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The tangent to the curve y=6√x at the point (4,12) meets the axes at A and B. Show that the distance AB may be written in the form k√13, and state the value of k.
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So the equation of the tangent to the curve at (4,12) is
Now put x = 0 and y = 0 to determine the points of intersection with the co-ordinate axes, then use Pythagoras theorem to find the length of AB.
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y = 6(x^1/2)
dy/dx= 3x^(-1/2) thats the slope or the gradient
dy/dx at x = 4 is = 3*4^(-1/2) = 3/2
therefore the equation of the tangent y-y1=m(x-x1)
y - 12 = 3/2( x - 4 )
2y - 24 = 3x - 12
2y = 3x + 12
when y=0 ,it cuts the x axis
3x + 12 =0
x= -4
so let point A(-4,0)
when x = 0 it cuts the y axis
2y = 12
y = 6
so let point B (0,6)
the distance between to points =√(4^2 + 6^2) = 52 = 4√13
which isin the form of k√13 where k = 4
i hope it makes you clear....sorry if there is any confusion with symbols cos i was not sure how to use them
but the answer and method are correct
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k is not equal to 4. k is equal to 2.
By the way, helping someone with a problem does not mean doing their homework for them! It just means giving them a guiding hand so that they can do their homework.
Last edited by JaneFairfax (2007-12-01 12:10:55)
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oh sorry...i am new to this group...so i just helped by doing the some...sorry
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thanks guys.. helped lots
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k is not equal to 4. k is equal to 2.
By the way, helping someone with a problem does not mean doing their homework for them! It just means giving them a guiding hand so that they can do their homework.
Yes, well some people have different philosophies when it comes to what 'helping' people means, and they should be respected... but I'll give your suggestions a go.
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