Area of a conic section - 2

The figure shows the $xz$ -plot looking along the $y$ -axis. The blue lines are the outline of the right cone and the red line is the cut $x+y=5$ . Since the cut is parallel to the edge of the cone, the cut is a parabola of the form $y^2 = kx$ . The cut has a $z$ -intercept of $5$ and it meets the side of the cone where $y=x+10$ (the left edge) and $y=5-x$ (red line) meet or $x+10 = 5-x \implies x = - \frac 52$ . This point is the vertex of the parabola or $(0,0)$ ; and the open end its at the $z$ -intercept. Since the length of the red line is $\left(5+\frac 52\right)\sqrt 2 = \frac {15}{\sqrt 2}$ . On the parabola plane the open end points are $\left(\frac {15}{\sqrt 2}, \pm y_1 \right)$ , where $y_1= \sqrt{\frac {15}{\sqrt 2}k}$ is the $y$ -coordinate of the top open end point. Since the base of the corn is a circle of radius 10, we can find $y_1 = \sqrt{10^2-5^2} = 5 \sqrt 3$ . Therefore $\sqrt{\frac {15}{\sqrt 2}k}=5\sqrt 3\implies k = 3\sqrt 2$ and the equation of the parabola is $y^2 = 3\sqrt 2 x$ . The area of the cut is then:

$\begin{aligned} A & = 2 \int_0^{\frac {15}{\sqrt 2}} y\ dx = 2 \int_0^{\frac {15}{\sqrt 2}} \sqrt{3\sqrt 2 x}\ dx \\ & = \frac 43 \sqrt{3\sqrt 2} x^\frac 32 \bigg|_0^{\frac {15}{\sqrt 2}} = \frac 43 \sqrt{3\sqrt 2} \cdot \frac {15}{\sqrt 2} \sqrt{\frac {15}{\sqrt 2}} \\ & = 10 \sqrt {150} = 50\sqrt 6 \end{aligned}$

Therefore $a+b=50+6=\boxed{56}$ .

The equation of the cone is $x^2 + y^2 - (z - 10)^2 = 0$

The cutting plane is spanned by the two mutually perpendicular unit vectors $u_1 = (0, 1, 0)$ and $u_2 = (-1, 0, 1)/\sqrt{2}$

therefore, points of this plane can be expressed in terms of $u_1$ and $u_2$ as

$p = (5, 0, 0) + x' u_1 + y' u_2 = (5 - y'/\sqrt{2}, x' , y'/\sqrt{2})$

Now substitute the above expression in the equation of the cone, you'd get,

$(5 - y'/\sqrt{2})^2 + x'^2 - (y'/\sqrt{2} - 10)^2 = 0$

Expand and simplify,

$25 - 5 \sqrt{2} y' + \frac{1}{2} y'^2 + x'^2 - \frac{1}{2} y'^2 + 10\sqrt{2} y' - 100 = 0$

So

$5\sqrt{2} y' + x'^2 - 75 = 0$

That is, $y' = 7.5 \sqrt{2} - \dfrac{\sqrt{2}}{10} x'^2$

The limits of $x'$ are the ones that make $y'$ equal to zero, and theses are $x' = \pm 5 \sqrt{3}$

The area is found by integration

$\text{Area} = \displaystyle \int_{-5\sqrt{3}}^{5 \sqrt{3}} 7.5 \sqrt{2} - \frac{\sqrt{2} }{10} x'^2 \hspace{4pt} dx' = 75 \sqrt{6} - \frac{\sqrt{2}}{30} ( (2)(125)(3)\sqrt{3} ) = 75 \sqrt{6} - \sqrt{2}( 25) \sqrt{3} = \boxed{50 \sqrt{6}}$