Question

find the equation of cone with vertex (alpha,beta,gama)and base y2-4ax=0,z=0.

Answer 1

Answer on Question #52828 – Math – Analytic Geometry

Find the equation of cone with vertex (alpha, beta, gama) and base $y^2 - 4ax = 0, z = 0$.

Solution

Any line through vertex $(\alpha, \beta, \gamma)$ is $\frac{x - \alpha}{l} = \frac{y - \beta}{m} = \frac{z - \gamma}{n}$.

Any point on this line is $(x, y, z) = (lr + \alpha, mr + \beta, nr + \gamma)$.

It will lie on the given parabola $y^2 - 4ax = 0, z = 0$ if, for the same value of $r$, we have

$$\left\{ \begin{array}{l} nr + \gamma = 0 \\ (mr + \beta)^2 - 4 \cdot a(lr + \alpha) = 0 \end{array} \right.$$

Then

$$\left\{ \begin{array}{l} r = -\gamma / n \\ (mr + \beta)^2 - 4 \cdot a(lr + \alpha) = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} r = -\gamma / n \\ \left(\beta - \gamma \frac{m}{n}\right)^2 - 4 \cdot a\left(\alpha - \gamma \frac{l}{n}\right) = 0 \end{array} \right.$$

So, eliminating $l, m, n$

$$\left(\beta - \gamma \frac{m}{n}\right)^2 - 4 \cdot a\left(\alpha - \gamma \frac{l}{n}\right) = 0 \Rightarrow (\beta n - \gamma m)^2 - 4 \cdot a n(\alpha n - \gamma l) = 0$$ $$\left(\beta (z - \gamma) - \gamma (y - \beta)\right)^2 - 4 \cdot a \alpha (z - \gamma)^2 + 4 a \gamma (z - \gamma) (x - \alpha) = 0$$

The equation of cone is $z^2 (\beta^2 - 4a\alpha) + 4a\gamma xz + 4a\alpha\gamma z - 2\beta\gamma yz - 4a\gamma^2 x + \gamma^2 y^2 = 0$.

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