When the coordinate axes are rotated by an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ about the origin, then…

1. $x^2-y^2=9$
2. $x^2+y^2+2 x y=4$
3. $x^2+y^2=9$
4. $x^2-y^2+9=0$

Solution

When the coordinate axis rotated by are angle $\theta$ about origin, then $ \begin{aligned} & \quad x=\cos \theta x_1-\sin \theta y_1 \text { and } \\ & y=\sin \theta x_1+\cos \theta y_1 \\ & \text { Here, } \theta=\tan ^{-1} \frac{3}{4} \end{aligned} $
$ \Rightarrow \tan \theta=\frac{3}{4} \Rightarrow \cos \theta=\frac{4}{5} \text { and } \sin \theta=\frac{3}{5} $ So, $\quad x=\frac{4}{5} x_1-\frac{3}{5} y_1$ and $y=\frac{3}{5} x_1+\frac{4}{5} y_1$ $x$ and $y$ satisfy the equation $ \begin{aligned} & x^2+y^2=9 \\ & \Rightarrow\left(\frac{4}{5} x_1-\frac{3}{5} y_1\right)^2+\left(\frac{3}{5} x_1+\frac{4}{5} y_1\right)^2=9 \end{aligned} $ $ \begin{aligned} & \Rightarrow \frac{16}{25} x_1^2+\frac{9}{25} y_1^2-\frac{24}{25} x_1 y_1 \\ & +\frac{9}{25} x_1^2+\frac{16}{25} y_1^2+\frac{24}{25} x_1 y_1=9 \\ & \Rightarrow x_1^2+y_1^2=9 \\ & \end{aligned} $ So, transformed equations is $x^2+y^2=9$

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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