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Section: Mathematics

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Question : 12 of 89

Marks: +1, -0

After the coordinate axes are rotated through an angle

\;\frac{\pi}{4}

in the anti clockwise direction without shifting the origin, if the equation

x^2+y^2-2x-4y-20=0

ax^2+2hxy+by^2+2gx+2fy+c=0

in the new coordinate system, then

\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}=

[AP EAPCET 23 May 2025 S2]

$\; -20$
$\; -25$
$\; -30$
$\; -35$

Solution:

\because x^2+y^2-2x-4y-20=0

After rotation of axes through

\theta=\frac{\pi}{4}

, it transforms to

ax^2+2hxy+by^2+2gx+2fy+c=0

in the given equation

a=1, h=0, b=1, g=-1, f=-2, c=-20

After rotation of

\theta=\frac{\pi}{4}

a' = a\cos^2\theta + b\sin^2\theta + 2h\cos\theta\sin\theta

\;\;\; = 1\left(\frac{1}{\sqrt{2}}\right)^2 + 1\left(\frac{1}{\sqrt{2}}\right)^2 + 0 = 1

b' = a\sin^2\theta + b\cos^2\theta - 2h\sin\theta\cos\theta

\; = 1\left(\frac{1}{\sqrt{2}}\right)^2 + 1\left(\frac{1}{\sqrt{2}}\right)^2 - 0 = 1

also,

h' = (b-a)\sin\theta\cos\theta + h(\cos^2\theta - \sin^2\theta)

\; = 0+0 = 0

Similarly,

x' = x\cos\theta - y\sin\theta = \frac{1}{\sqrt{2}}(x-y)

and

y' = x\sin\theta + y\sin\theta = \frac{1}{\sqrt{2}}(x+y)

then

{-2x'-4y'} = \frac{-2x+2y-4x-4y}{\sqrt{2}}

\; = \frac{-6x-2y}{\sqrt{2}}

\; = -3\sqrt{2}x - \sqrt{2}y = 2gx + 2fy

then

g = -\frac{3}{\sqrt{2}}, f = -\frac{1}{\sqrt{2}}, c = -20

So,

\;\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = \begin{vmatrix} 1 & 0 & -\frac{3}{\sqrt{2}} \\ 0 & 1 & -\frac{1}{\sqrt{2}} \\ -\frac{3}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & -20 \end{vmatrix}

\;\;\; = 1\left(-20-\frac{1}{2}\right) - \frac{3}{\sqrt{2}}\left(+\frac{3}{\sqrt{2}}\right)

\;\;\; = -\frac{41}{2} - \frac{9}{2} = -25

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