suprabeast

I have this problem, just wondered if anyone can help.

You may need a pen and paper

ok here goes:

y=ax+b is a tangent to (x^2)+(y^2)=64

prove that

(a^2)+1 = (b^2)/64

Can anyone clever do this?

suprabeast

oh and for those of you that dont know,

y=ax+b is a straight line

and

y^2+x^2=64 is a parabola!

dsmith

Dont have a proof right now, but I suspect its to with the gradients of the graphs being equal at the point of tangent.

So the differential of "X^2 + Y^2 = 64" will equal "a" and then substitute back in somewhere.

Unfortunately my calculus is more than a litte rusty

Deano

suprabeast

thanks mate, i'll try working with that!

Any other bods?

dharbige

y^2 + x^2 = 64 is not a parabola. It's a circle, radius 8, centred on (0,0).

The tangent crosses the x-axis at y=b, and crosses the x-axis at x=(-b/a). This gives the triangle [(-b/a, 0), (0,0), (0,b)]
The hypotinuse of this triangle is sqrt( b^2 + ((b/a)^2).
At the point the tangent touches the circle, it will meet a radius of the circle at 90degrees. This radius will, of course, pass through point (0,0).
We now have a couple of similar triangles, and it can be seen that:
sqrt( b^2 + ((b/a)^2) / b = (b/a) / 8

Squaring both sides gives:
(b^2 + (b^2 / a^2)) / b^2 = (b^2/a^2)/64

Factorizing the lhs gives:
1 + 1/a^2 = (b^2/a^2)/64

Multiplying both sides by a^2 gives:
a^2 + 1 = b^2/64

[Edited to correct a minus sign that should have been an equals sign]


[Edited by dharbige - 5/28/2003 1:42:24 PM]

dsmith

Nice

Completely missed it was a circle

Deano