Area of a conic section - 3

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 $y=\frac x2+5$ . Since the cut cuts through the whole cone from one edge to another, the conic section cut out is an ellipse of the form $\frac {x^2}{a^2} + \frac {y^2}{b^2}=1$ .

The cut has a $z$ -intercept of $-10$ and it meets the side of the cone where $y=10-x$ (the right edge) and $y=x\frac x2+5$ (red line) meet or $10-x = \frac x2+5 \implies x = \frac {10}3$ , where the $y$ -coordinate is $y_1 = \frac 12 \cdot \frac {10}3 + 5 = \frac {20}3$ and the length of the red line $\sqrt{\left(10+\frac {10}3\right)^2+\left(\frac {20}3\right)^2} = \frac {20\sqrt 5}3$ . Since the $\frac {20\sqrt 5}3$ red line is the longest line on the ellipse, we can take it as the $x$ -axis of the ellipse plane. Then the major axis $a$ is half the length of the red line or $a= \frac {10\sqrt 5}3$ .

The center of the ellipse is at the midpoint of the red line, where its $x$ -coordinate is $x_0=\frac {-10+\frac {10}3}2 = - \frac {10}3$ and its $z$ -coordinate is $z_0=\frac {x_0}2 + 5 = \frac {10}3$ . The radius of the cross-sectional circle of the right cone at height $z$ is given by $r=10-z$ . For $z_0$ , $r_0 = 10-\frac {10}3 = \frac {20}3$ . The minor axis $b=\sqrt{ r_0^2-x_0^2} = \sqrt{\left(\frac {20}3\right)^2-\left(\frac {10}3\right)^2} = \frac {10\sqrt 3}3$ .

Therefore the area of the ellipse is $\pi ab = \pi \cdot \frac {10\sqrt 5}3 \cdot \frac {10\sqrt 3}3 \approx \boxed{135.2}$ .

If we take the $xz$ plane section of the cone, we obtain a triangle, and our cutting plane is line $z = 5 + \frac{1}{2} x$ . Since the angle that the cutting plane makes with the base is less than the angle that the curved surface of the cone makes makes with the base, then our conic section is an ellipse. The eccentricity of this ellipse is given by the formula,

$e = \dfrac{ \sin \beta }{\sin \alpha}$

where $\beta = \tan^{-1} \frac{1}{2}$ , and $\alpha = \tan^{-1} \dfrac{h}{r}$

Now the major axis of the ellipse extends along the line $z = 5 + \frac{1}{2} x$ , and this line intersects the ellipse at $x_1 = -10$ (at the base) , and the other point is obtained by finding the intersection of the above line with $z= 10 - x$ which leads to $x_2 = \dfrac{10}{3}$ , hence the two vertices along the major axis of the ellipse are $(-10, 0, 0)$ and $(\frac{10}{3}, 0, \frac{20}{3} )$ and this means that the length of major axis

$2 a = \sqrt{ (10 + \frac{10}{3} )^2 + (\frac{20}{3})^2 } \Rightarrow a = \dfrac{10\sqrt{5}}{3} = 7.4536$

Now, the eccentricity is related to the lengths of semi-major axis $a$ and semi-minor axis length $b$ by, $e = \sqrt{1 - (\frac{b}{a})^2 } \Rightarrow b = a \sqrt{1 - e^2 }$ . Subtituting the value of $e$ obtained above gives

$b = 5.7735$

Thus, finally, the area of the ellipse is $\text{Area} = \pi a b = 135.2$