
Find the equation of axis of the given hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ which is equally inclined to the axes.
A. $y=x+1$
B. $y=x-1$
C. $y=x+2$
D. $y=x-2$
Answer
161.7k+ views
Hint: To solve this question we will first determine the value of ${{a}^{2}}$ and ${{b}^{2}}$by comparing the standard equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ with the given equation of hyperbola. To find the equation of axis which is equally incline to the axis we will use the equation of tangent as equation of tangent is equally inclined to the axis $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$. Substituting the values of $m,{{a}^{2}}$ and ${{b}^{2}}$ in the formula we will determine the equation.
Formula Used: Equation of tangent : $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$
Complete step by step solution: We have given an equation of hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ and we have to find the equation of axis of the given hyperbola which will be equally inclined to the axes.
We know that the standard form of the equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, so we will now compare it with the given equation of hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ and determine the value of ${{a}^{2}}$ and ${{b}^{2}}$.
$\begin{align}
& \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2} \\
& {{a}^{2}}=3 \\
& {{b}^{2}}=2
\end{align}$
Now we know that equation of tangent is equally inclined to the axis so to find the equation of given hyperbola which will be equally inclined to axes we will use the formula of equation of tangent.
$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$.
As the axes are equally inclined it means that the slope will be equal to $1$. We will now substitute $m=1,\,{{a}^{2}}=3$ and ${{b}^{2}}=2$in the formula $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$.
$\begin{align}
& y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}} \\
& y=1.x+\sqrt{3{{(1)}^{2}}-2} \\
& y=x+1
\end{align}$
The equation of axis of the given hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ which is equally inclined to the axes is
$y=x+1$.
Option ‘A’ is correct
Note: Equally inclined axes means that the angle between both the axis will be ${{45}^{0}}$.Substituting $\theta ={{45}^{0}}$ in the formula of slope $m=\tan \theta $ we will get the value of slope.
$\begin{align}
& m=\tan \theta \\
& m=\tan {{45}^{0}} \\
& m=1
\end{align}$
Hence slope of equally inclined axes is one.
Formula Used: Equation of tangent : $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$
Complete step by step solution: We have given an equation of hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ and we have to find the equation of axis of the given hyperbola which will be equally inclined to the axes.
We know that the standard form of the equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, so we will now compare it with the given equation of hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ and determine the value of ${{a}^{2}}$ and ${{b}^{2}}$.
$\begin{align}
& \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2} \\
& {{a}^{2}}=3 \\
& {{b}^{2}}=2
\end{align}$
Now we know that equation of tangent is equally inclined to the axis so to find the equation of given hyperbola which will be equally inclined to axes we will use the formula of equation of tangent.
$y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$.
As the axes are equally inclined it means that the slope will be equal to $1$. We will now substitute $m=1,\,{{a}^{2}}=3$ and ${{b}^{2}}=2$in the formula $y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$.
$\begin{align}
& y=mx+\sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}} \\
& y=1.x+\sqrt{3{{(1)}^{2}}-2} \\
& y=x+1
\end{align}$
The equation of axis of the given hyperbola $\dfrac{{{x}^{2}}}{3}-\dfrac{{{y}^{2}}}{2}=1$ which is equally inclined to the axes is
$y=x+1$.
Option ‘A’ is correct
Note: Equally inclined axes means that the angle between both the axis will be ${{45}^{0}}$.Substituting $\theta ={{45}^{0}}$ in the formula of slope $m=\tan \theta $ we will get the value of slope.
$\begin{align}
& m=\tan \theta \\
& m=\tan {{45}^{0}} \\
& m=1
\end{align}$
Hence slope of equally inclined axes is one.
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