Question

The transformed equation of $a{x}^2+bxy+c{y}^2=2$ is $2X^2+Y^2=1,$ when the axes are rotated through an angle of ${45}^{\circ }$ in anticlockwise direction. Then which of the following is (are) true?

• A. $a^2+b^2+c^2=20$
• B. Roots of the equation $a{m}^2+bm+c=0$ are real
• C. Arithmetic mean of $a$ and $c$ is $b+1$
• D. Range of the function $f\left(y\right)=\frac{ay-b}{by-c}$ is $R-\left\{\frac{3}{2}\right\}$

Answer 1

Answer: (C), (D)

Range of the function $f\left(y\right)=\frac{ay-b}{by-c}$ is $R-\left\{\frac{3}{2}\right\}$

New equation is $2X^2+Y^2=1$ and $\theta ={45}^{\circ }$
$X=x\cos \theta +y\sin \theta =\frac{x+y}{\sqrt{2}}$
$Y=-x\sin \theta +y\cos \theta =\frac{-x+y}{\sqrt{2}}$
$\therefore$ Original equation is $2{\left(\frac{x+y}{\sqrt{2}}\right)}^2+{\left(\frac{-x+y}{\sqrt{2}}\right)}^2=1$

$\Rightarrow 3x^2+2xy+3y^2=2$
$\therefore a=3,b=2,c=3$
$\Rightarrow a^2+b^2+c^2=9+4+9=22$

$a{m}^2+bm+c=0$
$\Rightarrow 3m^2+2m+3=0$
$D=4-36=-32<0$
$\Rightarrow$ Roots are imaginary.

A.M. of $a$ and $c$ is $\frac{a+c}{2}=3=b+1$

$f\left(y\right)=\frac{ay-b}{by-c}=\frac{3y-2}{2y-3}$
Domain is $R-\left\{\frac{3}{2}\right\}$
Let $\frac{3y-2}{2y-3}=z$
$\Rightarrow 3y-2=2yz-3z$
$\Rightarrow y=\frac{3z-2}{2z-3}$
$\therefore$ Range is $R-\left\{\frac{3}{2}\right\}$