Question

use \\( \theta = 30 ^ { \circ } \\) to write \\( \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 9 } = 1 \\) in the \\( x ^ { \prime } y ^ { \prime } \\) plane. then identify the conic. \\( 18 \left( x ^ { \prime } \
ight) ^ { 2 } + 16 \sqrt { 3 } x ^ { \prime } y ^ { \prime } - 2 \left( y ^ { \prime } \
ight) ^ { 2 } - 88 = 0 \\); ellipse \\( 23 \left( x ^ { \prime } \
ight) ^ { 2 } + 26 \sqrt { 3 } x ^ { \prime } y ^ { \prime } - 3 \left( y ^ { \prime } \
ight) ^ { 2 } - 144 = 0 \\); ellipse \\( 18 \left( x ^ { \prime } \
ight) ^ { 2 } + 16 \sqrt { 3 } x ^ { \prime } y ^ { \prime } - 2 \left( y ^ { \prime } \
ight) ^ { 2 } - 88 = 0 \\); hyperbola \\( 23 \left( x ^ { \prime } \
ight) ^ { 2 } + 26 \sqrt { 3 } x ^ { \prime } y ^ { \prime } - 3 \left( y ^ { \prime } \
ight) ^ { 2 } - 144 = 0 \\); hyperbola

Explanation:

Step1: Recall the rotation of axes formulas

The rotation of axes formulas are \(x = x'\cos\theta - y'\sin\theta\) and \(y=x'\sin\theta + y'\cos\theta\). Given \(\theta = 30^{\circ}\), \(\cos\theta=\frac{\sqrt{3}}{2}\) and \(\sin\theta=\frac{1}{2}\). So \(x=\frac{\sqrt{3}}{2}x'-\frac{1}{2}y'\) and \(y = \frac{1}{2}x'+\frac{\sqrt{3}}{2}y'\).

Step2: Substitute \(x\) and \(y\) into the given equation

Substitute \(x\) and \(y\) into \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=1\).

$$
LATEXBLOCK0
$$

Step3: Find a common denominator and simplify

The common denominator of \(16\) and \(36\) is \(144\).

$$
LATEXBLOCK1
$$

Step4: Identify the conic

For a general second - degree equation \(Ax^{2}+Bxy + Cy^{2}+Dx + Ey+F = 0\), the discriminant is \(B^{2}-4AC\). Here \(A = 23\), \(B=- 26\sqrt{3}\), \(C=-3\). Then \(B^{2}-4AC=(-26\sqrt{3})^{2}-4\times23\times(-3)=2028 + 276=2304>0\). When \(B^{2}-4AC>0\), the conic is a hyperbola.

Answer:

\(23(x')^{2}+26\sqrt{3}x'y'-3(y')^{2}-144 = 0\); hyperbola