• Algebra > Rotation of Conics

ROTATION OF CONIC SECTIONS

Most algebra textbooks give an equation of a rotated conic in the xy-plane as \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \) and rotate it to standard form in the x'y'-plane. We usually think of this as rotating the axes rather than rotating the conic. The presence of the B'xy term is an indication that the conic has been rotated. So the goal is to rotate it so B'xy term is eliminated.

Here, we will take a different approach. Instead of dealing with the x'y'-plane, we will simply assume all equations are in the xy-plane. We will rotate the conic and rotate it back, eliminating the use of the x' and y' notations. Therefore, we will use primes to designate a rotated conic in the xy-plane so as not to confuse these coefficients with the standard form of the conics.

Before we start rotation, let’s look at some parabolas and hyperbolas you may have already encountered that have a rotation angle.

The Parabola Hiding

The equation of the form \(\sqrt{x}+\sqrt{y}=\sqrt{c}\) is a portion of a parabola that has been rotated by 45°. I discovered this form when finding the tangent-generated curve discussed here. Since neither x, y, nor c can be negative, this parabola lies completely in the first quadrant. It has the line y = x as the line of symmetry. The endpoints of this portion (0, c) and (c, 0) is tangent to the two axes.

By isolating and squaring the radicals twice, the standard form of this parabola is revealed to be \(x^2-2xy+y^2-2cx-2cy+c^2=0\). Since x, y, and c are not constrained to positive numbers, we can see the full parabola. The images below show a comparison of the two graphed.

We know the equation \(x^2-2xy+y^2-2ax-2ay+a^2=0\) represents a parabola because the determinant for a rotated conic, \(B^2-4AC\), gives 4 – 4(1)(1) = 0. If the determinant is 0, the conic is a parabola. Note that in the rotated form, both the x² and y² terms can be present for an equation of a parabola. (Only parabolas opening up and down or left and right have either x² or y² term, but not both.

In general, equations of the form \(\sqrt{ax}+\sqrt{by}=\sqrt{c}\) represent a portion of the parabola. To get the full curve, one must take the negative of the terms also. Three equations complete the graph. The two remaining are: \(-\sqrt{ax}+\sqrt{by}=\sqrt{c}\) and \(\sqrt{ax}-\sqrt{by}=\sqrt{c}\). By squaring twice, we can convert this form into standard form: \(a^2x^2-2abxy+b^2y^2-2acx-2bcy+c^2=0\). The determinant of this is (2ab)² – 4a²b² = 0. And as we can see, the presence of the xy term indicated a rotated conic.

The Hyperbola Hiding

Equations of the form xy = a or \(y=\dfrac{a}{x}\) represent hyperbolas rotated by 45°. These are the simplest hyperbolas hiding in plain site.

Another set of equations that represent hyperbolas are rational functions with two asymptotes: one vertical and one slanted. They have the form: \(y=\dfrac{Ax^2+Dx+F}{Bx+E}\). Written in standard form, this becomes: \(Ax^2-Bxy+Dx-Ey+F=0\). There is no y² term. The absence of the y², which has C as the coefficient, means the determinant will always be greater than 0: B² – 4AC = B² – 0 = B². This will always be greater than 0 whether B is positive or negative. Hence, these are hyperbolas.

Note that: of the 3 coefficients of the numberator, A, D, and F, up to 2 of them can be 0 and the equation will be a hyperbola nonetheless. If B is 0, then the equation simply becomes a quadratic.

The first asymptote is at the line \(x=-\dfrac{E}{B}\). The second asymptote is the line \(y=\dfrac{A}{B}x+\dfrac{DB-AE}{B^2}\).

Looking at the equation of the asymptote, we can easily see that if A = 0, then the asymptote will be a horizontal line. Therefore, hyperbolas of the form \(y=\dfrac{Dx+F}{Bx+E}\) have a horizontal and a vertical asymptote, even if D = 0. The horizontal asymptote is the line \(y=\dfrac{D}{B}\). If D = 0, then the x-axis becomes the asymptote.

An example of a “slanted” hyperbola with the equation \(y=\dfrac{x^2+3x+8}{4x-5}\) is graphed below.

Later, we will find the major axis and minor axis of this form by finding the angle of rotation.

Rotation Formulas

Rotation of a Point

To rotate conics, we will first find formulas of a point rotated by θ°. In the image above, we want to rotate point A by θ° so it is at point B given by coordinates (x', y'). The angle point A created with the x-axis is φ°. After rotation, \(x' = r\cos(\phi + \theta)\) and \(y' = r\sin(\phi + \theta)\). After using the trigonometric identity of the sum of angles, we get:

Equations for Rotation of Conics

Most textbooks give equations of a rotated conic and give examples of how to find its equation in standard form. Here, we want to take a conic in standard orientation and rotate it counterclockwise.

Now, suppose we want to translate point P(x, y) by (h, k). So our new point P' is (x + h, y + k). However, if we want to translate a function, we replace all x’s with x – h and y’s with y – k. Using the same concept, the above rotation is for points. We have to actually rotate in the opposite direction of above to rotate the conic counterclockwise. So, our rotation equations are below.

Now, we can rotate the general form of the conic: \(Ax^2+Cy^2+Dx+Ey+F=0\). Note that there is no B term because the B is the coefficient of the xy term. When the xy term is present, this indicates the conic is already rotated. So, after rotation, we should end up with the xy term.

Substituting these values into this equation, we have:

(i) \(A(x\cos\theta+y\sin\theta)^2+\) \(C(-x\sin\theta+y\cos\theta)^2+\) \(D(x\cos\theta+y\sin\theta)+\) \(E(-x\sin\theta+y\cos\theta)+\) \(F=0\)

(ii) \(A[(cos\theta)^2x^2+\sin(2\theta)xy+\) \((sin\theta)^2y^2]+C[(\sin\theta)^2x^2-\sin(2\theta)xy+\) \((\cos\theta)^2y^2)]+\) \(D(\cos\theta)x + D(\sin\theta)y - \) \(E(\sin\theta)x + E(\cos\theta)y + F = 0\)

Collecting like term we have the final equation of a rotated conic by angle θ°:

Example 1

Let’s show that x² – y² = a has the equation of the form xy = a/2 after rotating by 45°. A = 1, C = –1, F = –a, and the rest of the variables are 0. Both sine and cosine of 45° are √2/2 and sine of (2×45)° is 1. We have:

(i) \(\left (1\cdot\dfrac{2}{4} - 1\cdot\dfrac{2}{4}\right)x^2 + \left(1\cdot1 + 1\cdot1\right)xy + \) \(\left(1\cdot\dfrac{2}{4} - 1\cdot\dfrac{2}{4}\right)y^2+(0)x + (0)y-a=0\)

This gives us \(2xy-a=0\) or \(xy = \dfrac{a}{2}\). The image below shows the two graphs where a = 2.

Note that when A = C, the xy term is 0. It would seem the rotation had no effect since they xy did not appear. Actually, the conic is a circle which may not be centered at the origin if D, E, and F are not 0 or different from the original values. Hence, a circle can still be rotated about the origin.

Example 2

Let’s rotate a parabola with the equation y = x² by 30°. This equation in general form is x² – y = 0. A = 1, C = 0, D = 0, E = –1, and F = 0. Also, sin 30° = 1/2, sin 60° = √3/2, and cos 30° = √3/2. Plugging all this in we get:

(i) \(\left(1\cdot \left(\dfrac{\sqrt{3}}{2}\right)^2 + 0\right)x^2 + (1 - 0)\dfrac{\sqrt{3}}{2}xy + \) \(\left(1\cdot \left(\dfrac{1}{2}\right)^2 + 0\right )y^2 +\) \(\left(0 + \dfrac{1}{2}\right)x - \left( 0+\dfrac{\sqrt{3}}{2}\right)y = 0\)

(ii) \(\dfrac{3}{4}x^2 + \dfrac{\sqrt{3}}{2}xy + \dfrac{1}{4}y^2 + \dfrac{1}{2}x - \dfrac{\sqrt{3}}{2}y = 0\)

The two parabolas are graphed below:

Rotating Conics to Standard Orientation

Conics with the xy term are rotated. We can rotate them back to their normal orientation so the equation is in standard form. This is accomplished by eliminating the xy term. We will take the general equation of the conic, apply the clockwise rotation formula and evaluate the variable(s) that will make the coefficient of the xy equal 0. The calculation, though not difficult, seems messy so we will skip some steps in between. Double-angle trigonometric identities will be used to get to the final result.

Our clockwise rotation formulas are: \(x = x'\cos\theta - y'\sin\theta\) and \(y = y'\cos\theta + x'\sin\theta\). We will drop the prime designation of variables.

(i) \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\)

(ii) \(A(x\cos\theta - y\sin\theta)^2 + \) \(B(x\cos\theta - y\sin\theta)(y\cos\theta + x\sin\theta) +\) \(C(y\cos\theta + x\sin\theta)^2 + D(x\cos\theta - y\sin\theta) + \) \( E(y\cos\theta + x\sin\theta) + F = 0\)

(iii) \(A[\cos^2\theta\cdot x^2 - \sin(2\theta)\cdot xy + \sin^2\theta\cdot y^2] + \) \( B[(\cos^2\theta - \sin^2\theta)xy +\) \((1/2)(\sin(2\theta))x^2 - (1/2)(\sin(2\theta))y^2] + \) \( C[\cos^2\theta\cdot y^2 + \sin(2\theta)\cdot xy +\) \(\sin^2\theta\cdot x^2] + \) \((D\cos\theta + E\sin\theta)x + \) \( (-D\sin\theta + E\cos\theta)y + F = 0\)

(iv) \(\left(A\cos^2\theta + \dfrac{1}{2}B\sin(2\theta)+C\sin^2\theta\right)x^2 + \) \( (-A\sin\(2\theta)+C\sin(2\theta) + B\cos(2\theta))xy +\) \(\left(A\sin^2\theta -\dfrac{1}{2}B\sin(2\theta) + C\cos^2\theta\right )y^2 + \) \( (D\cos\theta + E\sin\theta)x +\) \((-D\sin\theta + E\cos\theta)y + F = 0\)

Equation (iv) is the final result after rotation.

To rotate the above conic to its standard orientation, we need to know its angle of rotation. To find that, he have to eliminate the xy term by letting the coefficient of the xy term equal 0.

(i) \(-A\sin(2\theta)+C\sin(2\theta) + B\cos(2\theta) = 0\)

(ii) \((C-A)\sin(2\theta) + B\cos(2\theta) = 0\)

(iii) \(\dfrac{A-C}{B}=\cot 2\theta\)

(iv) \(2\theta = \cot^{-1}\dfrac{A-C}{B}\) or \(\theta = \dfrac{1}{2}\cot^{-1}\dfrac{A-C}{B}\)

The angle of rotation can also be stated using arctangent: \(\theta = \dfrac{1}{2}\tan^{-1}\dfrac{B}{C-A}\), however, we generally use the arccotangent.

The best way to rotate a conic is finding the rotation angle and then using the trigonometric formula above. However, we can eliminate the trigonometric calculations and formulate an equation of a standard orientation solely based on the coefficients. But first, let’s make sure we understand the angle of rotation, because the angle of rotation can turn out to be positive or negative.

The Angle of Rotation

We rotated the general conic in the clockwise direction and then found its angle of rotation. Therefore, a positive angle means clockwise rotation from rotated form to standard form, and a negative angle means counterclockwise rotation from rotated form to standard form. This may seem counterintuitive since we always associate a positive angle value as counterclockwise and negative angle value is clockwise. However, the convention is to rotate the axes rather than the conic itself. If we think of it as rotating the axes, then a positive value actually rotates the axes in the counterclockwise direction. Keepin this in mind, we move forward.

Knowing the above is critical because the trigonometric values are “sensitive” to the direction of angle measure. With that in mind, let’s calculate the trigonometric values needed to eliminate the trigonometric values from our rotation equation above. We will need to know the sine, cosine and twice the sine from the equation. We will look at two scenarios: one for a positive angle of rotation and one for a negative angle of rotation. The image below will help with this.

The range of arccotangent is –π/2 to π/2. Therefore, our rotation angles will fall in Quadrants I or IV. The image shows point A at angle 2θ, B at θ, C at –2θ and D at –θ.

If the rotation angle is positive, then sin(θ), cos(θ), and sin(2θ) are all positive since they fall in Quadrant I. If the rotation angle is negative, then cos(θ) is positive and sin(θ) and sin(2θ) are negative since they falls in Quadrant IV. We need to account for this when we evaluate the sine and cosine.

One more fact about the angle of rotation. The range of arccotangent gives us the following inequality: \(-\dfrac{\pi}{2} \leq \cot^{-1}\left(\dfrac{A-C}{B}\right) \leq \dfrac{\pi}{2}\). Our angle of rotation is half that. Therefore, we divide the whole inequality by 2 and get \(-\dfrac{\pi}{4} \leq \dfrac{1}{2}\cot^{-1}\left(\dfrac{A-C}{B}\right) \leq \dfrac{\pi}{4}\). Therefore, we never get an angle of rotation that is more than 45°. You may note that a conic can be rotated either clockwise or counterclockwise to render it in standard orientation. Since we used a clockwise rotation formula, the angle of rotation we obtain will rotate the conic clockwise unless θ is greater than 45°, i.e. 2θ is greater than 90°. If 2θ is greater than 90°, we will get a negative angle of rotation.

The Right Triangle

Now that we have the rotation angle figured out, let’s construct a right triangle with angle measure of 2θ and side |A – C| adjacent to the angle and side |B| that is opposite the angle. The third side will be \(\sqrt{(A-C)^2+B^2}\) using the Pythagorean theorem.

From this triangle, we can determine that \(\cos(2\theta) = \dfrac{|A-C|}{\sqrt{(A-C)^2+B^2}}\). Why the absolute value bars? Because cosine will never be negative in the first and fourth Quadrants according to our circle figure above. Therefore, we need to ensure the numberator is positive in case A < C.

Now, \(\sin(2\theta) = \pm\dfrac{|B|}{\sqrt{(A-C)^2+B^2}}\). If angle of rotation is negative, we need to take the negative value.

Using the half-angle identities, we have \(\cos\theta = \sqrt{\dfrac{1}{2}+\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}}\), which will always be positive. And \(\sin\theta = \pm\sqrt{\dfrac{1}{2}-\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}}\). If the angle of rotation is negative, we need to take the positive and if negative, we take the negative.

Calculation of the New Coefficients

Now that we have determined our trigonometric values, we are ready to substitute these values in our rotation equation above to eliminate the trigonometric values. Let’s determine the new coefficients one at a time. The first coefficient A' (for the x² term) is: \( \left(A\cos^2\theta + \dfrac{1}{2}B\sin(2\theta)+C\sin^2\theta\right) = \) \( A\left(\dfrac{1}{2}+\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}\right) + \) \(\dfrac{1}{2}B\cdot\dfrac{\pm |B|}{\sqrt{(A-C)^2+B^2}} + \) \( C\left(\dfrac{1}{2}-\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}} \right)\). This gives us:

(i) \(A' = \dfrac{A+C}{2} + \dfrac{(A-C)|A-C|\pm |B|\cdot B}{2\sqrt{(A-C)^2+B^2}}\)

If A > C, then equation (i) simplifies to \(A' = \dfrac{A+C}{2} + \dfrac{1}{2}\sqrt{(A-C)^2+B^2}\), because \(\pm |B|\cdot B\) becomes B² since the angle of rotation will have the same sign as B.

If C > A, then equation (i) simplifies to \(A' = \dfrac{A+C}{2} - \dfrac{1}{2}\sqrt{(A-C)^2+B^2}\), because \(\pm |B|\cdot B\) becomes –B² since the angle of rotation will have the opposite sign as B.

Without knowing which is greater, we cannot write \(\pm B\cdot B\) as \(\pm B^2\).

(ii) \(B' = 0\)

The xy term will be 0. The coefficient for the y² term was \( \left(A\sin^2\theta - \dfrac{1}{2}B\sin(2\theta)+C\cos^2\theta\right)\). The new coefficient will be (calculations are similar to above):

(iii) \(C' = \dfrac{A+C}{2} - \dfrac{(A-C)|A-C|\pm |B|\cdot B}{2\sqrt{(A-C)^2+B^2}}\)

If A > C, then equation (i) simplifies to \(A' = \dfrac{A+C}{2} - \dfrac{1}{2}\sqrt{(A-C)^2+B^2}\).

If C > A, then equation (i) simplifies to \(A' = \dfrac{A+C}{2} + \dfrac{1}{2}\sqrt{(A-C)^2+B^2}\).

From our equation, the coefficient of x was \(D' = (D\cos\theta + E\sin\theta)x\). We need to apply the change of sign here to sine if the angle of rotation is negative. The new D is:

(iv) \(D' = D\sqrt{\dfrac{1}{2}+\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}} \text{ }\pm \) \(E\sqrt{\dfrac{1}{2}-\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}}\).

If A = C, then the equation simplifies to \(D' = D\sqrt{\frac{1}{2}} \pm \) \(E\sqrt{\frac{1}{2}} = \frac{D \pm E}{\sqrt{2}} \).

The coefficient of y was \(E' = (E\cos\theta - D\sin\theta)y\). The sign of D is negative, so we need to reverse sign. Therefore, the new E is:

(v) \(E' = E\sqrt{\dfrac{1}{2}+\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}}\text{ } \mp \) \(D\sqrt{\dfrac{1}{2}-\dfrac{|A-C|}{2\sqrt{(A-C)^2+B^2}}}\).

If A = C, then the equation simplifies to \(E' = E\sqrt{\frac{1}{2}} \mp \) \(D\sqrt{\frac{1}{2}} = \frac{E \mp D}{\sqrt{2}} \).

In (v), because D was negative to begin with, if sine is negative, we take the positive sign and if sine is positive, we take the negative sign.

(vi) \(F' = F\)

Using these formulas based on the coefficients of the rotated conic eliminates finding the angle of rotation and takes trigonometric calculations out of the “equation”. However, you can see there are a lot of “ifs” that go along with this method. The foolproof method is still using trigonometry.

Example 3

Let’s try this on an equation: \(6x^2+4\sqrt{3}xy+2y^2-9x+9\sqrt{3}y-63 = 0\).

Angle of rotation: Let’s find the angle of rotation first: \(\theta = \dfrac{1}{2}\cot^{-1}\dfrac{6-2}{4\sqrt{3}} = \dfrac{1}{2}\cot^{-1}\dfrac{\sqrt{3}}{3}\). One may recognize that the arccotangent value is π/3. Therefore, the angle of rotation if π/6 or 30°.

The angle of rotation is positive and A is greater than C, so we can use the short formulas for A' and C'. Let’s do some number crunching starting with A'.

\(A'=\dfrac{6+2}{2}+ \dfrac{1}{2}\sqrt{(6-2)^2+(4\sqrt{3})^2} =\) \( 4+\dfrac{\sqrt{64}}{2}=8\)

\(C'=\dfrac{6+2}{2}-\dfrac{1}{2}\sqrt{(6-2)^2+(4\sqrt{3})^2} = \) \( 4-\dfrac{\sqrt{64}}{2}= 0\)

\(D'=-9\sqrt{\dfrac{1}{2}+\dfrac{4}{2\sqrt{64}}} + 9\sqrt{3}\sqrt{\dfrac{1}{2}-\dfrac{4}{2\sqrt{64}}} = \) \(-\dfrac{9\sqrt{3}}{2}+\dfrac{9\sqrt{3}}{2} = 0 \)

\(E'=9\sqrt{3}\sqrt{\dfrac{1}{2}+\dfrac{4}{2\sqrt{64}}} - (-9)\sqrt{\dfrac{1}{2}-\dfrac{4}{2\sqrt{64}}} = \) \(\dfrac{9\sqrt{3}\sqrt{3}}{2} + \dfrac{9}{2} = 18\)

F' remains –63. So the equation that will orient the parabola in standard form is: \(8x^2+18y-63=0\). Both conics are shown below in the image.

You can see that the positive angle of rotation meant the parabola rotated clockwise from rotated to standard form.

Example 4

Problem: Rotate the hyperbola we saw earlier represented as a rational function: \(y = \dfrac{3x^2+3x+8}{4x-5}\). Write its equation in standard form.

Solution: Let’s write the equation is general form first: \(3x^2 - 4xy + 3x +5y+8 = 0\).

Now, we need to find the angle of rotation to first to determine what signs to use. The angle of rotation is \(\theta = \dfrac{1}{2}\cot^{-1}\dfrac{3-0}{-4} = -26.56\). The angle of rotation is negative. Now, we can use our formulas to find the coefficients one by one using the bottom sign.

(i) \(A' = \dfrac{A}{2} + \dfrac{A|A| - |B|B}{2\sqrt{A^2+B^2}} = \dfrac{3}{2} + \dfrac{3\cdot3 - 4\cdot(-4)}{2\sqrt{9+16}} = \dfrac{3}{2} + \dfrac{5}{2} = 4\).

We applied the negative sign for the negative angle of rotation above.

(ii) \(C' = \dfrac{A}{2} - \dfrac{A|A| - |B|B}{2\sqrt{A^2+B^2}} = \dfrac{3}{2} - \dfrac{3\cdot3 - 4\cdot(-4)}{2\sqrt{9+16}} = \dfrac{3}{2} - \dfrac{5}{2} = -1\).

The only difference between A' and C' is substraction instead of addition.

(iii) \(D' = D\sqrt{\dfrac{1}{2}+\dfrac{|A|}{2\sqrt{(A)^2+B^2}}} \text{ } - \) \( E\sqrt{\dfrac{1}{2}-\dfrac{|A|}{2\sqrt{(A)^2+B^2}}} = \) \(3\sqrt{\dfrac{1}{2}}+\dfrac{3}{10} \text{ } - 5\sqrt{\dfrac{1}{2}}-\dfrac{3}{10} = \) \(6\sqrt{\dfrac{1}{5}} - 5\sqrt{\dfrac{1}{5}} = \dfrac{1}{\sqrt{5}}\).

The calculation for E' is similar to D' except the sign we use is positive.

(v) \(E' = 10\sqrt{\dfrac{1}{5}} + 3\sqrt{\dfrac{1}{5}} = \dfrac{13}{\sqrt{5}}\).

The constant coefficient F' remains the same. Therefore, our hyperbola equation after rotation is: \(4x^2 - y^2 + \dfrac{1}{\sqrt{5}}x + \dfrac{13}{\sqrt{5}}y + 8 = 0\). The graph below shows our hyperbolas rotated and the angle of rotation. It seems the hyperbola is symmetric to the y-axis after rotation. Actually, the axis of symmetry is very close to the y-axis.

To put this equation in standard form, we need to complete the squares of both x and y. After a minor rearrangement, our equation is \(4x^2 + \dfrac{1}{\sqrt{5}}x - y^2 + \dfrac{13}{\sqrt{5}}y + 8 = 0\).

(vi) \(4\left(x + \dfrac{1}{8\sqrt{5}}\right)^2 - 4\cdot\dfrac{1}{320} - \left(y - \dfrac{13}{2\sqrt{5}}\right)^2 + \dfrac{169}{20} + 8 = 0\)

(vii) \(4\left(x + \dfrac{1}{8\sqrt{5}}\right)^2 - \left(y - \dfrac{13}{2\sqrt{5}}\right)^2 = 4\cdot\dfrac{1}{320} - \dfrac{169}{20} - 8\)

(viii) \(4\left(x + \dfrac{1}{8\sqrt{5}}\right)^2 - \left(y - \dfrac{13}{2\sqrt{5}}\right)^2 = -\dfrac{263}{16}\)

(ix) \(-\dfrac{\left(x + \dfrac{1}{8\sqrt{5}}\right)^2}{\dfrac{263}{64}} + \dfrac{\left(y - \dfrac{13}{2\sqrt{5}}\right)^2}{\dfrac{263}{16}} = 1\)

Equation (ix) puts our equation in standard form and we have our center and the lengths of a and b.

I rotated the center of the standard hyperbola by the angle of rotation to find the center of the rotated hyperbola. This point is K in the image and it is located at \(\left(\dfrac{5}{4},\dfrac{21}{8}\right)\). Note that it seems like K falls on the segment connecting the origin and V₁ but it is not. The point K actually lies on the segment connecting the two brances of the hyperbola and it happens to be close to this segment.

To find the eccentricity, we will find c of the hyperbola first: \(c^2 = a^2 + b^2 = \dfrac{263}{16} + \dfrac{263}{4} = \dfrac{1315}{16}\). The eccentricity is \(e = \dfrac{c}{a} = \dfrac{\dfrac{\sqrt{1315}}{4}}{\dfrac{\sqrt{263}}{4}} = \sqrt{5} \approx 2.23\).

The Determinant

We can use the determinant value of the general form of the equation to quickly tell if a conic is a circle, ellipse, parabola, or hyperbola. Let’s take a look at the general equation in standard orientation first.

(a) \(Ax^2+Cy^2+Dx+Ey+F=0\)

We first notice that if A and C are both equal, we have a circle. Therefore, we need to first rule out the conic being a circle.

Next, we recall that a parabola has either the x² term or the y² term is missing, but not both. So either A or C is 0. In either case, AC = 0.

Next, we recall that the signs of x² term and the y² are the same for an ellipse. And in order for the conic to be an elllipse, A and C cannot be equal, unless you think of a circle as a special ellipse with the major and minor axes being equal. Therefore, AC is greater than 0.

That leaves the hyperbola. And it follows that if AC < 0, it is a hyperbola, because the signs of x² term and the y² are opposite.

In the rotated form, we can multiply A and C, expand, simpify, and investigate the result with the criteria above to determin which conic it is. The A and C are multiplied here with some steps skipped for the final result.

(i) \(A'C' = \left(\dfrac{A+C}{2} + \dfrac{(A-C)|A-C|\pm B^2}{2\sqrt{(A-C)^2+B^2}}\right)\left(\dfrac{A+C}{2} - \dfrac{(A-C)|A-C|\pm B^2}{2\sqrt{(A-C)^2+B^2}}\right)\)

This is like a difference of squares multiplication and the middle term cancels out, making it somewhat easier. Note that, when multiplied out, the quantity \((A-C)^2|A-C|^2\) equals \((A-C)^4\), so we can remove the absolute value bars and treat is as one now. And \(\pm B^2 \cdot \pm B^2 = B^2\), so we can remove the plus/minus sign since that quantity will always be positive now. This greatly simplies to the following:

(ii) \(A'C'=\dfrac{(A+C)^2}{4}-\dfrac{[(A-C)^2+B^2]^2}{4[(A-C)^2+B^2]}\)

(iii) \(A'C'=\dfrac{(A+C)^2}{4}-\dfrac{(A-C)^2+B^2}{4}\)

(iv) \(A'C'= \dfrac{A^2+2AC+C^2-A^2+2AC-C^2-B^2}{4}\)

(v) \(A'C' = \dfrac{1}{4}(-B^2+4AC)\)

Since we are comparing the product AC to 0, we can remove the 1/4 factor and call our determinant –B² + 4AC (not to be confused with the eccentricity, e, or the determinant of the quadratic equation). It is certainly amazing that all of that reduced to a simple product. When comparing, we can change this to the inverse inequality so that the term is B² – 4AC.

If B² – 4AC = 0, the conic is a parabola.

If B² – 4AC < 0, the conic is an ellipse.

If B² – 4AC > 0, the conic is a hyperbola.

And we have shown how to easily determine the conic based on a simple formula.

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