Question

The transformed equation of \[ 4x^2-4xy+7y^2-24=0 \] in the new coordinate system when the axes are rotated through an angle \(\theta\) about the origin in the positive direction is \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \] If \[ 0<\theta<\frac{\pi}{4}, \] then \[ 1+b^2\sin^2\theta= \]

• A. \(a^2\cos^2\theta\)
• B. \(a^2+\cos^2\theta\)
• C. \(a^2\sin^2\theta\)
• D. \(\dfrac{a^2}{\sin^2\theta}\)

Hint:

For second-degree equations, first eliminate the \(xy\)-term using \[ \boxed{\tan2\theta=\frac{2H}{A-B}}, \] then diagonalize the quadratic form to identify the semi-axes of the conic.

Answer 1

The Correct Option is C

Step 1: Find the angle of rotation.& nbsp;

For the general second-degree equation

\[ Ax^2+2Hxy+By^2=0, \]

the angle of rotation satisfies

\[ \tan2\theta=\frac{2H}{A-B}. \]

Here,

\[ A=4,\qquad 2H=-4,\qquad B=7. \]

Hence,

\[ \tan2\theta = \frac{-4}{4-7} = \frac{4}{3}. \]

Since

\[ 0<\theta<\frac{\pi}{4}, \]

we obtain

\[ \sin2\theta=\frac{4}{5}, \qquad \cos2\theta=\frac{3}{5}. \]

Therefore,

\[ \sin^2\theta = \frac{1-\cos2\theta}{2} = \frac{1}{5}. \]

Step 2: Find \(a^2\) and \(b^2\).

The eigenvalues of

\[ \begin{pmatrix} 4 & amp; -2\\ -2 & amp; 7 \end{pmatrix} \]

are

\[ 3,\;8. \]

Thus, the transformed equation is

\[ 3X^2+8Y^2=24, \]

or equivalently,

\[ \frac{X^2}{8}+\frac{Y^2}{3}=1. \]

Hence,

\[ a^2=8, \qquad b^2=3. \]

Step 3: Evaluate the required expression.

Now,

\[ 1+b^2\sin^2\theta = 1+3\left(\frac{1}{5}\right) = \frac{8}{5}. \]

Also,

\[ a^2\sin^2\theta = 8\left(\frac{1}{5}\right) = \frac{8}{5}. \]

Therefore,

\[ 1+b^2\sin^2\theta = a^2\sin^2\theta. \]

Hence,

\[ \boxed{a^2\sin^2\theta}. \]

Thus, the correct option is

\[ \boxed{(C)}. \]