Question

The only integral surface of the equation and 2q (z - px- qy) 1+ q^2 which is circumscribed about paraboloid 2x= y^2 + z^2 and which touches it along its section by the plane y + 1 = 0 is = ?

Answer 1

$2q (z - px- qy) =1+ q^2$

$$f(x, y, z, p, q)=2q (z - px- qy) -1- q^2=0$$ $$f_x=-2pq, f_y=-2q^2, f_z=2q, f_p=-2qx, f_q=2(z-px)-4qy-2q$$

$$\dfrac{dx}{f_p}=\dfrac{dy}{f_q}=\dfrac{dz}{pf_p+qf_q}=\dfrac{dp}{-(f_x+pf_z)}=\dfrac{dq}{-(f_y+qf_z)}$$

$\dfrac{dx}{-2qx}=\dfrac{dy}{2(z-px)-4qy-2q}=\dfrac{dz}{-2pq+q(2(z-px)-4qy-2q)}=$

$=\dfrac{dp}{2pq-2pq}=\dfrac{dq}{2q^2-2q^2}$

$dp=dq=0$

$p=a,q=b$

$dz=pdx+qdy$

$z=ax+by+c$

by condition $2x= y^2 + z^2$ :

$2x= y^2 + (ax+by+c)^2$

since $y+1=0$ :

$2x= 1 + (ax-b+c)^2$