Conic Sections: questions with answers

With $\beta = 20^\circ$ and $\alpha = 30^\circ$, the plane is steeper than a generator. A plane parallel to a generator cuts one nappe in a parabola, which runs away without end; a plane steeper than a generator cuts through both nappes, giving a hyperbola in two branches. A plane that makes a larger angle with the axis than the generator, $\alpha < \beta \le 90^\circ$, cuts one nappe in a closed ellipse or circle.

The centre goes inside the brackets and the point on the circle only fixes the radius. From $C(1, -3)$ to $P(13, 2)$ the horizontal gap is 12 and the vertical gap is 5, so $r = 13$ and $r^2 = 169$. The equation is $(x - 1)^2 + (y + 3)^2 = 169$.

$(x + p)^2 = x^2 + 2px + p^2$, so the number inside the bracket is HALF the coefficient of $x$. Here that coefficient is 14, so the bracket entry is $14 \div 2 = 7$, giving $(x + 7)^2$. The number 49 is what gets ADDED, not what stands inside the bracket.

The standard form subtracts the centre, so $(x + 4)^2$ must be read as $\{x - (-4)\}^2$, giving $h = -4$, and $(y - 3)^2$ gives $k = 3$. The centre is $(-4, 3)$; the numbers printed inside the brackets are the NEGATIVES of the centre's coordinates.

From $\frac{2b^2}{a} = 128/15$ we get $b^2 = 64$ and $b = 8$. Solving the exchanged formula $\frac{2a^2}{b} = 128/15$ instead gives 3375/64. The value 64/15 is half the chord and 256/225 divides the chord by the wrong length - neither is a semi-axis.

$(-3, 0)$ lies on the directrix, but it is not the foot of the perpendicular from $P$. The y-coordinate of $P$ is 6, not 0, and a horizontal perpendicular keeps that y-coordinate. Measuring to $(-3, 0)$ therefore gives a slanting length, longer than the true distance 6. The perpendicular foot on the vertical line $x = -3$ is $(-3, 6)$, and the distance is $3 + 3 = 6$. The focus $(3, 0)$ is not on $x = -3$, so it cannot be the perpendicular foot on the directrix. For a point on this parabola, the distance to the focus does equal the distance to the directrix numerically, but the focus is a different point and does not locate the foot on the line.

The latus rectum is a chord PERPENDICULAR to the axis of the parabola, drawn through the focus, with both ends on the curve. Here the axis is the $x$-axis and the focus is $(3, 0)$, so the latus rectum is vertical. A segment lying ALONG the axis is not a chord of this kind: the $x$-axis meets the parabola only at the vertex.

The latus rectum is the WHOLE chord through a focus perpendicular to the major axis, and it runs from the curve on one side of the focus to the curve on the other. Each half measures $\frac{b^2}{a} = 225/17$, so the whole chord is $\frac{2b^2}{a} = 450/17$. Reporting 225/17 counts only the piece from the focus to the curve.

The vertices are the ends of the MAJOR axis, which here runs along the $y$-axis. A stated axis LENGTH is twice the semi-axis, so $a = 26 \div 2 = 13$ and the vertices are $(0, \pm 13)$. The points $(\pm 5, 0)$ are the ends of the minor axis, and $(\pm 13, 0)$ puts the right length on the wrong axis.

A stated axis LENGTH is $2a$, twice the semi-axis, so $a = 8 \div 2 = 4$ and $a^2 = 16$. Then $b^2 = a^2 - c^2 = 16 - 9 = 7$. Using 8 itself as $a$ gives the far larger denominators 64 and 55, an ellipse twice the size of the one described.

Here the focal distance 4 is larger than the semi-minor axis 3, so dividing by 3 gives 4/3, a value greater than 1. No ellipse can have that, because its foci lie inside it and $c < a$. Measuring against the semi-major axis instead gives $e = \frac{c}{a} = 4/5$.

The foci always lie on the MAJOR axis. The larger denominator 841 is under $y^2$, so the major axis is the $y$-axis and the foci are $(0, \pm c)$. With $a = 29$ and $b = 20$, $c^2 = 841 - 400$, so $c = 21$ and the foci are $(0, \pm 21)$. Putting them on the $x$-axis assumes the major axis is always horizontal.

An $x^2$ term means the axis of symmetry is the $y$-axis. Replacing $x$ by $-x$ in $x^2 = 16y$ leaves it unchanged, so reflecting in the $y$-axis maps the curve onto itself; replacing $y$ by $-y$ does not, so the $x$-axis is not an axis of symmetry.

A constant DIFFERENCE of the two distances gives a hyperbola; a constant SUM gives an ellipse. Here the sum is fixed at 16, which is greater than the gap 12, so a real curve exists and it is an ellipse with those two points as foci and major axis 16.

With the origin at the top, a foot of the arch is 10 m sideways - half of 20 - and 12 m DOWN, so its $y$-coordinate is negative: $(10, -12)$. A $y$-coordinate of $0$ would mean the foot is level with the origin, which is what happens only if the origin is put at the base instead of at the vertex.