Problem in understanding the result of finding the equation of a tangent plane to a central conicoid.

Question

I was studying about conicoids. There was a topic about finding the tangent plane to a conicoid (general). It went on like this:

Let the equation of the conicoid be $ax^2+by^2+cz^2=1.$ Any line through the point $(\alpha,\beta,\gamma)$ is of the form $\frac {x-\alpha}{l}=\frac {y-\beta}{m}=\frac {z-\gamma}{n}=r.$ The coordinates of any point on this line is $(\alpha+lr, \beta+mr,\gamma+nr).$ Hence, the intersection of the line and the conicoid , we have $$\begin{multline} a(\alpha+lr)^2+b(\beta+mr)^2+c(\gamma+nr)^2=1\\ \implies\\ r^2(al^2+bm^2+cn^2)+2r(al\alpha+bm\beta+cn\gamma)+(a\alpha^2+b\beta^2+c\gamma^2-1)=0.\end{multline}$$ If the line touches the conicoid then we have both the roots of this equation $0.$ Since $(\alpha, \beta,\gamma)$ lies on $ax^2+by^2+cz^2=1$, so we have, $a\alpha^2+b\beta^2+c\gamma^2=1.$ Hence, one root of $$r^2(al^2+bm^2+cn^2)+2r(al\alpha+bm\beta+cn\gamma)+(a\alpha^2+b\beta^2+c\gamma^2-1)=0$$ is zero and the other root is zero if $al\alpha+bm\beta+cn\gamma=0$ , which is the condition that the line is a tangent. Now, eliminating $(l,m,n)$ between $\frac {x-\alpha}{l}=\frac {y-\beta}{m}=\frac {z-\gamma}{n}=r.$ and $al\alpha+bm\beta+cn\gamma=0$ we get, the equation of the tangent plane as $ax\alpha+by\beta+cz\gamma=1.$

But my question is, I dont get how do they conclude $ax\alpha+by\beta+cz\gamma=1$ as the equation of the tangent plane? Of course, that's a equation of a plane but how is it the tangent plane? An infinite number of planes can pass through a single line. I am not quite getting the reason behind concluding this as the tangent plane, when this plane found out in the quoted text can be any plane such that the tangent to the conicoid at the point passes through the plane?

Answer 1

Alternatively, using the tangent cone concept: Expand knowing $a\alpha^2+b\beta^2+c\gamma^2-1=0$ $$a(x+\alpha)^2+b(y+\beta)^2+c(z+\gamma)^2-1=0$$ to translate the point to the origin to get $$ax^2+by^2+cz^2+2a\alpha x+2b\beta y+2c\gamma z+a\alpha^2+b\beta^2+c\gamma^2-1=0$$ So taking the tangent cone (cut the higher than linear terms) and translating back $$2a\alpha (x-\alpha)+2b\beta (y-\beta)+2c\gamma (z-\gamma)=0$$ or $$2(a \alpha x+ b \beta y +c\gamma z - ( a \alpha^2+b \beta^2+ c \gamma^2-1+1))=0$$ or $$2(c\gamma z + b \beta y + a \alpha x -1)=0.$$

Answer 2

Put $f(x,y,z)=ax^2+by^2+cz^2$, for $(x,y,z)\in\Bbb{R}^3$, and let $\mathscr{Q}$ the quadric defined by the equation $f(x,y,z)=1$. The plane tangent to $\mathscr{Q}$ at $(\alpha,\beta,\gamma)$ is orthogonal to the vector $\mathop{\overrightarrow{\rm grad}}\limits_{(\alpha,\beta,\gamma)}f=(2a\alpha,2b\beta,2c\gamma)$ and thus has a cartesian equation of the form $2a\alpha x+2b\beta y+2c\gamma z=\lambda$, for some $\lambda\in\mathbb{R}$.

Since this tangent plane passes through the point $(\alpha,\beta,\gamma)$ we have $2a\alpha^2+2b\beta^2+2c\gamma^2=\lambda$, hence $\lambda=2$ and the (halved) equation above turns out to be $a\alpha x+b\beta y+c\gamma z=1.$