Question

Show that the general equation of a cone which touches the co-ordinate plane is
a²x² + b²y² + c²z² - 2bcyz - 2cazx - 2abxy = 0​

Answer 1

Given : To prove : The general equation of a cone which touches the co-ordinate plane is

Let the plane touches the cone at (α,β,γ).

We know that the equation of the tangent plane to the cone $a^2x^2+b^2y^2+c^2z^2-2bcyz-2cazx-2abxy=0$  at the point (α,β,γ) is

We will proceed it by taking the contradiction to it

$x(a^2\alpha +ac\beta +ab\gamma)+y(bc\alpha +b\beta +ab\gamma)+z(bc\alpha +ca\beta +c\gamma)=0$

Using the contradictory equation of the cone is $x^2+2y^2-z^2+2yz-5zx+3xy=0.$

Here a=1,b=2,c=−3,f=1,g=−52,h=32.

The equation of the tangent plane is x(α+32β−52γ)+y(32α+2β+γ)+z(−52α+β−3γ)=0

The given plane is parallel to the generator whose direction ratios are 1,1,1 or whose direction cosines are 13√,13√,13√

So 13√(α+32β−52γ)+13√(32α+2β+γ)+13√(−52α+β−3γ)=0

9β2−9γ2=0

β=γ

$Hence this cannot be possible for a cone to draw two or more tangent at same point$

$Hence our contradiction is wrong.$

$\therefore$ The general equation of a cone which touches the co-ordinate plane is

$a^2x^2+b^2y^2+c^2z^2-2bcyz-2cazx-2abxy=0$

                                                                          Proved

Answer 2

→ For 3) you can proceed as follows:

→ (a² + b² + c²). (x² + y² + z² )= (ax+by+cz)² , because 36x25=900=30²  

→ Therefore,

→ a²x² + a²y² + a²z²+b²x²+b²y²+b²z²+c²x²+c²y²+c²z²=a²x²+b²y²+c²z²+2abxy+2acxz+2bcyz

→ or,   a²y² + a²z²+b²x²+b²z²+c²x²+c²y²-2abxy-2acxz-2bcyz=0

→ or, (ay-bx)²+(az-cx)²+(bz-cy)²=0

→ Therefore, if ay-bx, az-cx, bz-cy are real numbers, we get

→ ay-bx=0, az-cx=0 and bz-cy=0

→ Again assuming none of x,y,z is zero, we get

→ a/x=b/y, a/x=c/z and b/y=c/z

→ i.e , a/x=b/y=c/z= k say

→ Therefore a=kx, b=ky,c=kz

→ Therfore, a² + b² + c²=k²(x² + y² + z²)

→ Hence, 36=k².25

→ or, k²=36/25

→ or k=+6/5 or -6/5

→ Hence, (a+b+c) / (x+y+z)=(kx+ky+kz)/ (x+y+z)=k= +6/5 or -6/5