Show that the vertex lies on the surface $z^2(\frac{x}{a}+\frac{y}{b})=4(x^2+y^2)$ Two cones with a common vertex pass through the curves $z^2=4ax,y=0$ and $z^2=4by,x=0.$ The plane $z=0$ meets them in two conics which intersect in four concyclic points.Show that the vertex lies on the surface $z^2(\frac{x}{a}+\frac{y}{b})=4(x^2+y^2)$

Let the common vertex of the two cones is $(\alpha,\beta,\gamma)$ and let the direction cosines of the generator line of first cone which passes through $z^2=4ax,y=0$ be $l_1,m_1,n_1$ and the direction cosines of the generator line of second cone which passes through $z^2=4ay,x=0$ be $l_2,m_2,n_2.$
The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$The equation of the first cone is $(\gamma-\frac{z\beta}{y})^2=4a(\alpha-\frac{\beta x}{y})$
The generator line of the second cone is $\frac{x-\alpha}{l_2}=\frac{y-\beta}{m_2}=\frac{z-\gamma}{n_2}.$The equation of the second cone is $(\gamma-\frac{z\alpha}{x})^2=4b(\beta-\frac{\alpha y}{x})$
I am stuck here.Please help me.


The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$

Setting $y=0$ gives
$$\frac{x-\alpha}{l_1}=\frac{0-\beta}{m_1}=\frac{z-\gamma}{n_1}\implies x=-\beta\frac{l_1}{m_1}+\alpha,\quad z=-\beta\frac{n_1}{m_1}+\gamma\tag1$$
Also, 
$$\frac{l_1}{m_1}=\frac{x-\alpha}{y-\beta},\quad \frac{n_1}{m_1}=\frac{z-\gamma}{y-\beta}\tag2$$
From $(1)(2)$,
$$z^2=4ax\implies \left(-\beta\cdot \frac{z-\gamma}{y-\beta}+\gamma\right)^2=4a\left(-\beta\cdot\frac{x-\alpha}{y-\beta}+\alpha\right)\tag3$$

The generator line of the second cone is $\frac{x-\alpha}{l_2}=\frac{y-\beta}{m_2}=\frac{z-\gamma}{n_2}.$

Setting $x=0$ gives
$$\frac{0-\alpha}{l_2}=\frac{y-\beta}{m_2}=\frac{z-\gamma}{n_2}\implies y=-\alpha\frac{m_2}{l_2}+\beta,\quad z=-\alpha\frac{n_2}{l_2}+\gamma\tag4$$
Also,
$$\frac{m_2}{l_2}=\frac{y-\beta}{x-\alpha},\quad \frac{n_2}{l_2}=\frac{z-\gamma}{x-\alpha}\tag5$$
From $(4)(5)$,
$$z^2=4by\implies \left(-\alpha\cdot \frac{z-\gamma}{x-\alpha}+\gamma\right)^2=4b\left(-\alpha\cdot \frac{y-\beta}{x-\alpha}+\beta\right)\tag6$$
Setting $z=0$ in $(3)(6)$ give
$$\left(-\beta\cdot \frac{0-\gamma}{y-\beta}+\gamma\right)^2=4a\left(-\beta\cdot\frac{x-\alpha}{y-\beta}+\alpha\right)\implies y^2\gamma^2=4a(\alpha y-\beta x)(y-\beta)$$
$$\implies b\alpha (\gamma^2-4a\alpha) y^2+4ab\alpha \beta xy+4ab\alpha^2\beta y-4ab\alpha \beta^2 x=0\tag7$$
$$\left(-\alpha\cdot \frac{0-\gamma}{x-\alpha}+\gamma\right)^2=4b\left(-\alpha\cdot \frac{y-\beta}{x-\alpha}+\beta\right)\implies x^2\gamma^2=4b(x\beta-\alpha y)(x-\alpha)$$
$$\implies a\beta (\gamma^2-4b\beta) x^2+4ab\alpha\beta xy+4a\beta^2 b\alpha x-4ba\beta \alpha^2 y=0\tag8$$
The four intersection points of $(7)$ with $(8)$ satisfy
$$b\alpha (\gamma^2-4a\alpha) y^2+4ab\alpha \beta xy+4ab\alpha^2\beta y-4ab\alpha \beta^2 x+k(a\beta (\gamma^2-4b\beta) x^2+4ab\alpha\beta xy+4a\beta^2 b\alpha x-4ba\beta \alpha^2 y)=0$$
Since it represents a circle for some $k$, we have to have $k=-1$ and 
$$b\alpha (\gamma^2-4a\alpha)=-a\beta (\gamma^2-4b\beta) ,$$
i.e.
$$\gamma^2\left(\frac{\alpha}{a}+\frac{\beta}{b}\right)=4(\alpha^2+\beta^2)$$
