Question

Find the shortest and the longest distances from the point $(1,2,-1)$ to the sphere $x^2+y^2{z}^2=24$

Answer 1

$(1,2,-1)$

$x^2+y^2+z^2=24$

length = $\sqrt{1^2+2^2+(-1{)}^2}=\sqrt{1+4+1}=\sqrt{6}$

Now, $F(x,y,z,\lambda )=(x-1{)}^2+(y-2{)}^2+(z+1{)}^2-\lambda (x^2+y^2+z^2-24)$

$\Rightarrow x^2+1-2x+y^2+4-4y+z^2+1+2z-\lambda x^2-\lambda y^2-24\lambda$

$=(1+\lambda )x^2+(1-\lambda )y^2-2x+1-4y+4+2z+1+(1-\lambda )z^2+24\lambda$

$\Rightarrow \frac{\partial F}{\partial x}=2x+2\lambda x-2=0$ __(1)

$\frac{\partial F}{\partial y}=2y-2\lambda y-4=0$ ___(2)

$\frac{\partial F}{\partial z}=2z-2\lambda z+2=0$ ___(3)

$\frac{\partial F}{\partial \lambda }=-x^2-y^2+2y-z^2=0$ ___(4)

$x+\lambda x-1=0$

$x=\frac{1}{1+\lambda }$

$y-\lambda y-2=0$

$y=\frac{+2}{1-\lambda }$

$z+\lambda z+1=0$

$z=-\frac{1}{1-\lambda }$

we get ,

$y=-2z,x=\frac{-y}{2(1-y)}$

Now, putting x ^ z values in equation (4) we get

$-{(-\frac{y}{2(1-y)})}^2-y^2-{\left(\frac{-y}{2}\right)}^2+24=0$

$\frac{y^2}{4(1-y{)}^2}+y^2+\frac{y^2}{4}-24=0$

$y^2+4y^2(1-y{)}^2+y^2(1-y{)}^2-24\times 4(1-y{)}^2=0$

$y^2+4y^2(1+y^2-2y)+y^2(1+y^2-2y)-96(1+y^2-2y)=0$

$y^2+4y^2+4y^4-8y^3+y^2+y^4-2y^3-96-96y^2+192y=0$

$5y^4-90y^2-10y^3+192y-96=0$