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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

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tangents and normals with exponentials and logs..............

Can anyone do this question?

The tangent to the curve y = ln x at the point (e, e) meets the x-axis at A and the y-axis at B.
a) Find the equation of this tangent.
b)Hence find the distance AB.

I can do part a) fine, I get the answer of y = 2x - e which is right.

But I don't know where to start with part b?

Any help would be much appreciated..............:smile:
Reply 1
Avatar for Gaz031
Gaz031
15
Raman87
Can anyone do this question?

The tangent to the curve y = ln x at the point (e, e) meets the x-axis at A and the y-axis at B.
a) Find the equation of this tangent.
b)Hence find the distance AB.

I can do part a) fine, I get the answer of y = 2x - e which is right.

But I don't know where to start with part b?

Any help would be much appreciated..............:smile:


Calculate the coordinates of A and B, by substituting y=0 and x=0 respectively into the equation for the tangent.
Reply 2
Avatar for Raman87
Raman87
OP
ohhhhhhhhhhhhh yeah!!! why didn't i think of that?!?! :redface:

It all seems so clear now...............:smile:

Thank you!! :biggrin:
Reply 3
Avatar for Raman87
Raman87
OP
wait a sec, i still havent got the answer :frown:

I have found A to be (1/2e, 0) and B to be (0, -e)

So using pythag..... distance AB appears to be root (e^2 + 1/4e^2)

The answer is supposed to be (root5e) / 2

Any help would be much appreciated......:smile:
Raman87
wait a sec, i still havent got the answer :frown:

I have found A to be (1/2e, 0) and B to be (0, -e)

So using pythag..... distance AB appears to be root (e^2 + 1/4e^2)

The answer is supposed to be (root5e) / 2

Any help would be much appreciated......:smile:

AB^2=(e/2-0)^2+(0-(-e))^2
=e^2/4+e^2
=5e^2/4
hence
AB=rt(5)e/2
Reply 5
Avatar for Raman87
Raman87
OP
thanks!!

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