Answer to Question #217499 in Analytic Geometry for Rishu

Question

find the equation of cone whose vertex is(5, 4,3) and base curve 3x2+2y2=6, y+z=0.

Answer 1

The vertex $V(5,4,3)$. Let $a,b,c$ be the direction ratios of generator of the cone.

Then the equations of the generator are

$$\frac{x-5}{a}=\frac{y-4}{b}=\frac{z-3}{c}=t$$

The coordinates of any points on the generator are $(5+at,4+bt,3+ct)$.

For some $t\in \R, (5+at,4+bt,3+ct)$ lies on the guiding curve.

Therefore $3(5+at)^2+2(4+bt)^2=6$ and $4+bt+3+ct=0$ .

Thus, $t=-\dfrac{7}{b+c}$.

From this we get

$$3(5-\dfrac{7a}{b+c})^2+2(4-\dfrac{7b}{b+c})^2=6$$

$$3(5b+5c-7a)^2+2(4b+4c-7b)^2=6(b+c)^2$$

$$75b^2+75c^2+147a^2+150bc-210ab-210ac$$

$$+18b^2-48bc+32c^2-6b^2-12bc-6c^2=0$$

$$147a^2+87b^2+101c^2-210ab-210ac+90bc=0$$

$$147(x-5)^2+87(y-4)^2+101(z-3)^2$$

$$-210(x-5)(y-4)-210(x-5)(z-3)$$

$$+90(y-4)(z-3)=0$$

$$147x^2-1470x+3675+87y^2-696y+1392$$

$$+101z^2-606z+909-210xy+840x+1050y$$

$$-4200-210xz+630x+1050z-3150+90yz$$

$$-270y-360z+1080=0$$

The equation of cone is

$$147x^2+87y^2+101z^2-210xy-210xz+90yz$$

$$+84y+84z-294=0$$