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Reduce the equation to standard form and determine the nature of the conicoid: x2+y2+z2-yz-zx-xy-3x-6y-9z+21=0.
asked Jan 8 by big boss (190 points)  
edited Jan 12 by randomisation
Please suggest book for analytical geometry.
P N Chatterjee OR Shanti Narayan OR Krishna Series. I've read Shanti Narayan, nut others generally recommend the first one.
Complete book list here: http://www.upsc.xyz/book-list/mathematics-optional

1 Answer

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Here, $a=b=c=1, f=g=h=-\frac{1}{2}, u=-\frac{3}{2}, v=-3, w=-\frac{9}{2}, d=21.$

The discriminating cubic is

    $\begin{vmatrix}
a-\lambda & h & g\\
h & b-\lambda & f\\
g & f & c-\lambda
\end{vmatrix} = 0$

or  \begin{vmatrix}
1-\lambda & -\frac{1}{2} & -\frac{1}{2}\\
-\frac{1}{2} & 1-\lambda & -\frac{1}{2}\\
-\frac{1}{2} & -\frac{1}{2} & 1-\lambda
\end{vmatrix} = 0

or $\lambda^3-3\lambda^2+\frac{9}{4}\lambda = 0$

or $\lambda(2\lambda-3)^2=0$

or $\lambda = 0,\frac{3}{2},\frac{3}{2}$

As two roots of the discriminating cubic are equal and third root is zero. So it is either a paraboloid of revolution or right circular cylinder.

The direction ratios of the axis are given by

$al + hm + gn =0, hl + bm + fn = 0, gl + fm + cn = 0$

$\implies l-m/2-n/2 =0, -l/2 + m - n/2, -l/2-m/2+n=0 \\ \implies . 2l - m - n = 0, -l + 2m - n = 0, - I - m + 2n =0$

Solving, we get, $l=m=n=\frac{1}{\sqrt{3}}$

Now $k = uI + vm + wn$

or $k = (- 3/2) (1/\sqrt{3}) + (- 3) (1/\sqrt{3}) + (- 9/2) ( 1/\sqrt{3}) = - 3\sqrt{3} \neq 0$

Therefore, the reduced equation is $\lambda_1 x^2 + \lambda_2 y^2 + 2kz=0$

or $\frac{3}{2}x^2 + \frac{3}{2} y^2 + 2(-3\sqrt{3})z = 0$

or $x^2 = y^2 = 4\sqrt{3}z$, which represents a paraboloid of revolution.‚Äč
answered Jan 8 by randomisation (1,920 points)  
edited Jan 12 by randomisation
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