Find the lengths of transverse axis and conjugate axis, eccentricity, the coordinates of focus, vertices, length of the latus-rectum and equation of the directrices of the following hyperbola $16{{x}^{2}}-9{{y}^{2}}=144$?
Answer
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Hint: We start solving the problem by converting the given equation of hyperbola into the standard form. We then find the value of ${{a}^{2}}$, ${{b}^{2}}$ after converting to standard form. We then make use of the fact that length of transverse axis, length of conjugate of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is 2a, 2b to proceed through the problem. We then make use of the fact that eccentricity of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $e=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}$ to proceed further through the problem. We then make use the fact that the coordinates of foci, coordinates of vertices of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\left( \pm ae,0 \right)$, $\left( \pm a,0 \right)$ to proceed further through the problem. We then make use of fact that the length of latus-rectum of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{2{{b}^{2}}}{a}$ and the equation of directrices of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $x=\pm \dfrac{a}{e}$ to complete the problem.
Complete step by step answer:
According to the problem, we are asked to find the lengths of transverse axis and conjugate axis, eccentricity, the coordinates of focus, vertices, length of the latus-rectum and equation of the directrices of the following hyperbola $16{{x}^{2}}-9{{y}^{2}}=144$.
We have given the equation of hyperbola as $16{{x}^{2}}-9{{y}^{2}}=144$.
We know that the standard form of the hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. Let us convert the given equation to this form.
So, we have $\dfrac{16{{x}^{2}}}{144}-\dfrac{9{{y}^{2}}}{144}=1$.
$\Rightarrow \dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{16}=1$. Comparing this with standard form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we get ${{a}^{2}}=9\Leftrightarrow a=3$, ${{b}^{2}}=16\Leftrightarrow b=4$.
We know that the length of transverse axis of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is 2a. So, the length of the transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $2\left( 3 \right)=6$.
We know that the length of conjugate axis of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is 2b. So, the length of the transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $2\left( 4 \right)=8$.
We know that the eccentricity of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $e=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $e=\sqrt{\dfrac{9+16}{9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}$.
We know that the coordinates of foci of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\left( \pm ae,0 \right)$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is \[\left( \pm \left( 3\times \dfrac{5}{3} \right),0 \right)=\left( \pm 5,0 \right)\].
We know that the coordinates of vertices of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\left( \pm a,0 \right)$. So, the length of the transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is \[\left( \pm 3,0 \right)\].
We know that the length of latus-rectum of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{2{{b}^{2}}}{a}$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is \[\dfrac{2\times 16}{3}=\dfrac{32}{3}\].
We know that the equation of directrices of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $x=\pm \dfrac{a}{e}$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $x=\pm \dfrac{3}{\dfrac{5}{3}}\Leftrightarrow x=\pm \dfrac{9}{5}$.
Note:
Whenever we get this type of problems, we first try to convert them to the standard form to make the process simpler. We should not confuse the standard notations of the hyperbola while solving this problem. We should not make calculation mistakes while solving this problem. Similarly, we can expect problems to find the similar properties of the hyperbola $9{{y}^{2}}-16{{x}^{2}}=144$.
Complete step by step answer:
According to the problem, we are asked to find the lengths of transverse axis and conjugate axis, eccentricity, the coordinates of focus, vertices, length of the latus-rectum and equation of the directrices of the following hyperbola $16{{x}^{2}}-9{{y}^{2}}=144$.
We have given the equation of hyperbola as $16{{x}^{2}}-9{{y}^{2}}=144$.
We know that the standard form of the hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. Let us convert the given equation to this form.
So, we have $\dfrac{16{{x}^{2}}}{144}-\dfrac{9{{y}^{2}}}{144}=1$.
$\Rightarrow \dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{16}=1$. Comparing this with standard form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we get ${{a}^{2}}=9\Leftrightarrow a=3$, ${{b}^{2}}=16\Leftrightarrow b=4$.
We know that the length of transverse axis of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is 2a. So, the length of the transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $2\left( 3 \right)=6$.
We know that the length of conjugate axis of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is 2b. So, the length of the transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $2\left( 4 \right)=8$.
We know that the eccentricity of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $e=\sqrt{\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $e=\sqrt{\dfrac{9+16}{9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}$.
We know that the coordinates of foci of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\left( \pm ae,0 \right)$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is \[\left( \pm \left( 3\times \dfrac{5}{3} \right),0 \right)=\left( \pm 5,0 \right)\].
We know that the coordinates of vertices of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\left( \pm a,0 \right)$. So, the length of the transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is \[\left( \pm 3,0 \right)\].
We know that the length of latus-rectum of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $\dfrac{2{{b}^{2}}}{a}$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is \[\dfrac{2\times 16}{3}=\dfrac{32}{3}\].
We know that the equation of directrices of $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ is $x=\pm \dfrac{a}{e}$. So, the length of transverse axis of $16{{x}^{2}}-9{{y}^{2}}=144$ is $x=\pm \dfrac{3}{\dfrac{5}{3}}\Leftrightarrow x=\pm \dfrac{9}{5}$.
Note:
Whenever we get this type of problems, we first try to convert them to the standard form to make the process simpler. We should not confuse the standard notations of the hyperbola while solving this problem. We should not make calculation mistakes while solving this problem. Similarly, we can expect problems to find the similar properties of the hyperbola $9{{y}^{2}}-16{{x}^{2}}=144$.
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