Find a Tangent Plane Parallel to x+2y+3z=1 on the Curve y=x2+z2

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I'm completely lost on this question, and it's due tomorrow morning. Help?

Homework Statement


What point on y=x2+z2 is the tangent plane parallel to the plane x+2y+3z=1?

Homework Equations


y=x2+z2
x+2y+3z=1

The Attempt at a Solution


I have no idea what to do...

Thanks!
 
Last edited:
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The gradient of the function F(x,y,z) is always perpendicular to the surface F(x,y,z)= constant at the point (x,y,z). Now, do you know how to find the equation of a plane, given a normal vector and one point on that plane?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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