Show that locus of a point from which three mutually perpendicular tangent lines can be drawn to paraboloid $x^2+y^2+2z=0$ is $x^2+y^2+4z=1$.
Attempt: Assumed the point to be $(\alpha,\beta,\gamma)$ and formed equation of 3 lines passing from $(\alpha,\beta,\gamma)$ and with DR's $(l_1,m_1,n_1)$,$(l_2,m_2,n_2)$ and $(l_3,m_3,n_3)$ and as the lines are mutually perpendicular the properties $l_1l_2+m_1m_2+n_1n_2=0$ etc holds true.
But I cannot move further in the direction to solve the problem. Is the approach incorrect or needs to be followed with some changes? Any help appreciated
For any three prependicular line $l_1^2 + l_2^2 + l_3^2 =1$ , $m_1^2+m_2^2+m_3^2 =1$ and $n_1^2+n_1^2+n_1^2 =1 , l_1m_1 + l_2m_2+l_3m_3 = 0 $ and similarly for $\sum_{i=1}^3m_in_i = 0$ $\sum_{i=1}^3l_in_i = 0 $.
Now let line to be $\frac{x-\alpha}{l_1} = \frac{y-\beta}{m_1} = \frac{z-\gamma}{n_1}$ solve it with parabola and make root equal using $b^2-4ac= 0$ which will result into an equation like $(l_1\alpha + m_1\beta + n_1\gamma)^2 = (\alpha^2 + \beta^2 + 2\gamma)(l_1^2+m_1^2)$
There will be two more similar equation in $l_2,m_2,n_2$ and $l_3,m_3,n_3$. Add these three equations and use the above mentioned relations. Hope it helps.