NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120684
The equation of the hyperbola whose foci are \((6,5),(-4,5)\) and eccentricity \(5 / 4\) is
1 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1\)
2 \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
3 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1\)
4 None of these
Explanation:
A Centre of the hyperbola is the mid point of the line joining the two foci, therefore, the coordinates of the centre are \((1,5)\). Now distance between the foci \(=10\)
\(\Rightarrow 2 \mathrm{ae}=10 \Rightarrow \mathrm{ae}=5 \Rightarrow \mathrm{a}=4 \quad[\because \mathrm{e}=5 / 4]\)
Now, \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \Rightarrow \mathrm{b}=3\)
Hence, the equation of the hyperbola is
\(\frac{(\mathrm{x}-1)^2}{16}-\frac{(\mathrm{y}-5)^2}{9}=1 \text {. }\)
COMEDK-2018
Hyperbola
120686
The foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). Find the equation of the hyperbola if its eccentricity is 2 .
1 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
2 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
3 \(\frac{x^2}{3}-\frac{y^2}{4}=1\)
4 \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
Explanation:
B The equation of the ellipse is \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\therefore \mathrm{a}^2=25\) and \(\mathrm{b}^2=9\)
Eccentricity of ellipse \(=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
So the co-ordinates of the foci are \(( \pm 4,0)\)
\(\therefore\) the foci of hyperbola are \(( \pm 4,0)\)
Let, \(e^{\prime}\) be the eccentricity of the required hyperbola and its equation be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^{12}}\)
The coordinates of the foci of the hyperbola are \(\left( \pm \mathrm{ae}^{\prime}\right.\), 0 ) or \(( \pm 4,0)\)
It is given that \(\mathrm{e}^{\prime}=2\)
therefore, \(\mathrm{a}^{\prime} \times 2=4\) i.e \(\mathrm{a}^{\prime}=2\) or \(\mathrm{a}^{\mathrm{a}^2}=4\)
\(\mathrm{b}^{\prime 2}=\mathrm{a}^{\prime 2}\left(\mathrm{e}^{1^2}-1\right)\)
\(=2^2\left(2^2-1\right)\)
\(=4 \times 3\)
\(=12\)
Therefore, the required equatio of the hyperbola is
\(\frac{x^2}{4}-\frac{y^2}{12}=1\)
COMEDK-2018
Hyperbola
120687
The point (at \(\left.{ }^2, 2 b t\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) for
1 all values of \(t\)
2 \(\mathrm{t}^2=2+\sqrt{5}\)
3 \(\mathrm{t}^2=2-\sqrt{5}\)
4 No real values of \(t\)
Explanation:
B
The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right) \text { lies on the hyperbola }\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore \frac{\left(\mathrm{at}^2\right)^2}{\mathrm{a}^2}-\frac{(2 \mathrm{bt})^2}{\mathrm{~b}^2}=1 \Rightarrow \mathrm{t}^4-4 \mathrm{t}^2-1=0\)
\(\Rightarrow \mathrm{t}^4-4 \mathrm{t}^2+4-5=0\)
\(\Rightarrow\left(\mathrm{t}^2-2\right)^2=5 \Rightarrow \mathrm{t}^2-2= \pm \sqrt{5} \Rightarrow \mathrm{t}^2=2 \pm \sqrt{5}\)
\(\text { But } \mathrm{t}^2=2-\sqrt{5} \text { does not give real value. }\)
\(\text { So, } \mathrm{t}^2=2+\sqrt{5}\)
Ans: b
Exp:(B) The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).
But \(\mathrm{t}^2=2-\sqrt{5}\) does not give real value.
So, \(\quad \mathrm{t}^2=2+\sqrt{5}\)
COMEDK-2019
Hyperbola
120688
Find the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13 .
1 \(25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
2 \(144 \mathrm{x}^2-25 \mathrm{y}^2=900\)
3 \(5 \mathrm{x}^2-12 \mathrm{y}^2=30\)
4 \(12 x^2-5 y^2=30\)
Explanation:
A Let 2a and 2b be the transverse and conjugate axes and \(e\) be the eccentricity. The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
We have, \(2 \mathrm{~b}=5\) and \(2 \mathrm{ae}=13\)
Now, \(b^2=a^2\left(e^2-1\right) \Rightarrow b^2=a^2 e^2-a^2\)
\(\Rightarrow \frac{25}{4}=\frac{169}{4}-\mathrm{a}^2 \Rightarrow \mathrm{a}^2=\frac{144}{4} \Rightarrow \mathrm{a}^2=36\)
Thus, the equation of the hyperbola is
\(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{25 / 4}=1 \Rightarrow 25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
120684
The equation of the hyperbola whose foci are \((6,5),(-4,5)\) and eccentricity \(5 / 4\) is
1 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1\)
2 \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
3 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1\)
4 None of these
Explanation:
A Centre of the hyperbola is the mid point of the line joining the two foci, therefore, the coordinates of the centre are \((1,5)\). Now distance between the foci \(=10\)
\(\Rightarrow 2 \mathrm{ae}=10 \Rightarrow \mathrm{ae}=5 \Rightarrow \mathrm{a}=4 \quad[\because \mathrm{e}=5 / 4]\)
Now, \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \Rightarrow \mathrm{b}=3\)
Hence, the equation of the hyperbola is
\(\frac{(\mathrm{x}-1)^2}{16}-\frac{(\mathrm{y}-5)^2}{9}=1 \text {. }\)
COMEDK-2018
Hyperbola
120686
The foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). Find the equation of the hyperbola if its eccentricity is 2 .
1 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
2 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
3 \(\frac{x^2}{3}-\frac{y^2}{4}=1\)
4 \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
Explanation:
B The equation of the ellipse is \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\therefore \mathrm{a}^2=25\) and \(\mathrm{b}^2=9\)
Eccentricity of ellipse \(=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
So the co-ordinates of the foci are \(( \pm 4,0)\)
\(\therefore\) the foci of hyperbola are \(( \pm 4,0)\)
Let, \(e^{\prime}\) be the eccentricity of the required hyperbola and its equation be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^{12}}\)
The coordinates of the foci of the hyperbola are \(\left( \pm \mathrm{ae}^{\prime}\right.\), 0 ) or \(( \pm 4,0)\)
It is given that \(\mathrm{e}^{\prime}=2\)
therefore, \(\mathrm{a}^{\prime} \times 2=4\) i.e \(\mathrm{a}^{\prime}=2\) or \(\mathrm{a}^{\mathrm{a}^2}=4\)
\(\mathrm{b}^{\prime 2}=\mathrm{a}^{\prime 2}\left(\mathrm{e}^{1^2}-1\right)\)
\(=2^2\left(2^2-1\right)\)
\(=4 \times 3\)
\(=12\)
Therefore, the required equatio of the hyperbola is
\(\frac{x^2}{4}-\frac{y^2}{12}=1\)
COMEDK-2018
Hyperbola
120687
The point (at \(\left.{ }^2, 2 b t\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) for
1 all values of \(t\)
2 \(\mathrm{t}^2=2+\sqrt{5}\)
3 \(\mathrm{t}^2=2-\sqrt{5}\)
4 No real values of \(t\)
Explanation:
B
The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right) \text { lies on the hyperbola }\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore \frac{\left(\mathrm{at}^2\right)^2}{\mathrm{a}^2}-\frac{(2 \mathrm{bt})^2}{\mathrm{~b}^2}=1 \Rightarrow \mathrm{t}^4-4 \mathrm{t}^2-1=0\)
\(\Rightarrow \mathrm{t}^4-4 \mathrm{t}^2+4-5=0\)
\(\Rightarrow\left(\mathrm{t}^2-2\right)^2=5 \Rightarrow \mathrm{t}^2-2= \pm \sqrt{5} \Rightarrow \mathrm{t}^2=2 \pm \sqrt{5}\)
\(\text { But } \mathrm{t}^2=2-\sqrt{5} \text { does not give real value. }\)
\(\text { So, } \mathrm{t}^2=2+\sqrt{5}\)
Ans: b
Exp:(B) The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).
But \(\mathrm{t}^2=2-\sqrt{5}\) does not give real value.
So, \(\quad \mathrm{t}^2=2+\sqrt{5}\)
COMEDK-2019
Hyperbola
120688
Find the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13 .
1 \(25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
2 \(144 \mathrm{x}^2-25 \mathrm{y}^2=900\)
3 \(5 \mathrm{x}^2-12 \mathrm{y}^2=30\)
4 \(12 x^2-5 y^2=30\)
Explanation:
A Let 2a and 2b be the transverse and conjugate axes and \(e\) be the eccentricity. The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
We have, \(2 \mathrm{~b}=5\) and \(2 \mathrm{ae}=13\)
Now, \(b^2=a^2\left(e^2-1\right) \Rightarrow b^2=a^2 e^2-a^2\)
\(\Rightarrow \frac{25}{4}=\frac{169}{4}-\mathrm{a}^2 \Rightarrow \mathrm{a}^2=\frac{144}{4} \Rightarrow \mathrm{a}^2=36\)
Thus, the equation of the hyperbola is
\(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{25 / 4}=1 \Rightarrow 25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
120684
The equation of the hyperbola whose foci are \((6,5),(-4,5)\) and eccentricity \(5 / 4\) is
1 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1\)
2 \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
3 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1\)
4 None of these
Explanation:
A Centre of the hyperbola is the mid point of the line joining the two foci, therefore, the coordinates of the centre are \((1,5)\). Now distance between the foci \(=10\)
\(\Rightarrow 2 \mathrm{ae}=10 \Rightarrow \mathrm{ae}=5 \Rightarrow \mathrm{a}=4 \quad[\because \mathrm{e}=5 / 4]\)
Now, \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \Rightarrow \mathrm{b}=3\)
Hence, the equation of the hyperbola is
\(\frac{(\mathrm{x}-1)^2}{16}-\frac{(\mathrm{y}-5)^2}{9}=1 \text {. }\)
COMEDK-2018
Hyperbola
120686
The foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). Find the equation of the hyperbola if its eccentricity is 2 .
1 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
2 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
3 \(\frac{x^2}{3}-\frac{y^2}{4}=1\)
4 \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
Explanation:
B The equation of the ellipse is \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\therefore \mathrm{a}^2=25\) and \(\mathrm{b}^2=9\)
Eccentricity of ellipse \(=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
So the co-ordinates of the foci are \(( \pm 4,0)\)
\(\therefore\) the foci of hyperbola are \(( \pm 4,0)\)
Let, \(e^{\prime}\) be the eccentricity of the required hyperbola and its equation be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^{12}}\)
The coordinates of the foci of the hyperbola are \(\left( \pm \mathrm{ae}^{\prime}\right.\), 0 ) or \(( \pm 4,0)\)
It is given that \(\mathrm{e}^{\prime}=2\)
therefore, \(\mathrm{a}^{\prime} \times 2=4\) i.e \(\mathrm{a}^{\prime}=2\) or \(\mathrm{a}^{\mathrm{a}^2}=4\)
\(\mathrm{b}^{\prime 2}=\mathrm{a}^{\prime 2}\left(\mathrm{e}^{1^2}-1\right)\)
\(=2^2\left(2^2-1\right)\)
\(=4 \times 3\)
\(=12\)
Therefore, the required equatio of the hyperbola is
\(\frac{x^2}{4}-\frac{y^2}{12}=1\)
COMEDK-2018
Hyperbola
120687
The point (at \(\left.{ }^2, 2 b t\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) for
1 all values of \(t\)
2 \(\mathrm{t}^2=2+\sqrt{5}\)
3 \(\mathrm{t}^2=2-\sqrt{5}\)
4 No real values of \(t\)
Explanation:
B
The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right) \text { lies on the hyperbola }\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore \frac{\left(\mathrm{at}^2\right)^2}{\mathrm{a}^2}-\frac{(2 \mathrm{bt})^2}{\mathrm{~b}^2}=1 \Rightarrow \mathrm{t}^4-4 \mathrm{t}^2-1=0\)
\(\Rightarrow \mathrm{t}^4-4 \mathrm{t}^2+4-5=0\)
\(\Rightarrow\left(\mathrm{t}^2-2\right)^2=5 \Rightarrow \mathrm{t}^2-2= \pm \sqrt{5} \Rightarrow \mathrm{t}^2=2 \pm \sqrt{5}\)
\(\text { But } \mathrm{t}^2=2-\sqrt{5} \text { does not give real value. }\)
\(\text { So, } \mathrm{t}^2=2+\sqrt{5}\)
Ans: b
Exp:(B) The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).
But \(\mathrm{t}^2=2-\sqrt{5}\) does not give real value.
So, \(\quad \mathrm{t}^2=2+\sqrt{5}\)
COMEDK-2019
Hyperbola
120688
Find the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13 .
1 \(25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
2 \(144 \mathrm{x}^2-25 \mathrm{y}^2=900\)
3 \(5 \mathrm{x}^2-12 \mathrm{y}^2=30\)
4 \(12 x^2-5 y^2=30\)
Explanation:
A Let 2a and 2b be the transverse and conjugate axes and \(e\) be the eccentricity. The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
We have, \(2 \mathrm{~b}=5\) and \(2 \mathrm{ae}=13\)
Now, \(b^2=a^2\left(e^2-1\right) \Rightarrow b^2=a^2 e^2-a^2\)
\(\Rightarrow \frac{25}{4}=\frac{169}{4}-\mathrm{a}^2 \Rightarrow \mathrm{a}^2=\frac{144}{4} \Rightarrow \mathrm{a}^2=36\)
Thus, the equation of the hyperbola is
\(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{25 / 4}=1 \Rightarrow 25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hyperbola
120684
The equation of the hyperbola whose foci are \((6,5),(-4,5)\) and eccentricity \(5 / 4\) is
1 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1\)
2 \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
3 \(\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1\)
4 None of these
Explanation:
A Centre of the hyperbola is the mid point of the line joining the two foci, therefore, the coordinates of the centre are \((1,5)\). Now distance between the foci \(=10\)
\(\Rightarrow 2 \mathrm{ae}=10 \Rightarrow \mathrm{ae}=5 \Rightarrow \mathrm{a}=4 \quad[\because \mathrm{e}=5 / 4]\)
Now, \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \Rightarrow \mathrm{b}=3\)
Hence, the equation of the hyperbola is
\(\frac{(\mathrm{x}-1)^2}{16}-\frac{(\mathrm{y}-5)^2}{9}=1 \text {. }\)
COMEDK-2018
Hyperbola
120686
The foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). Find the equation of the hyperbola if its eccentricity is 2 .
1 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
2 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
3 \(\frac{x^2}{3}-\frac{y^2}{4}=1\)
4 \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
Explanation:
B The equation of the ellipse is \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{9}=1\)
\(\therefore \mathrm{a}^2=25\) and \(\mathrm{b}^2=9\)
Eccentricity of ellipse \(=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\)
So the co-ordinates of the foci are \(( \pm 4,0)\)
\(\therefore\) the foci of hyperbola are \(( \pm 4,0)\)
Let, \(e^{\prime}\) be the eccentricity of the required hyperbola and its equation be
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^{12}}\)
The coordinates of the foci of the hyperbola are \(\left( \pm \mathrm{ae}^{\prime}\right.\), 0 ) or \(( \pm 4,0)\)
It is given that \(\mathrm{e}^{\prime}=2\)
therefore, \(\mathrm{a}^{\prime} \times 2=4\) i.e \(\mathrm{a}^{\prime}=2\) or \(\mathrm{a}^{\mathrm{a}^2}=4\)
\(\mathrm{b}^{\prime 2}=\mathrm{a}^{\prime 2}\left(\mathrm{e}^{1^2}-1\right)\)
\(=2^2\left(2^2-1\right)\)
\(=4 \times 3\)
\(=12\)
Therefore, the required equatio of the hyperbola is
\(\frac{x^2}{4}-\frac{y^2}{12}=1\)
COMEDK-2018
Hyperbola
120687
The point (at \(\left.{ }^2, 2 b t\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) for
1 all values of \(t\)
2 \(\mathrm{t}^2=2+\sqrt{5}\)
3 \(\mathrm{t}^2=2-\sqrt{5}\)
4 No real values of \(t\)
Explanation:
B
The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right) \text { lies on the hyperbola }\)
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\therefore \frac{\left(\mathrm{at}^2\right)^2}{\mathrm{a}^2}-\frac{(2 \mathrm{bt})^2}{\mathrm{~b}^2}=1 \Rightarrow \mathrm{t}^4-4 \mathrm{t}^2-1=0\)
\(\Rightarrow \mathrm{t}^4-4 \mathrm{t}^2+4-5=0\)
\(\Rightarrow\left(\mathrm{t}^2-2\right)^2=5 \Rightarrow \mathrm{t}^2-2= \pm \sqrt{5} \Rightarrow \mathrm{t}^2=2 \pm \sqrt{5}\)
\(\text { But } \mathrm{t}^2=2-\sqrt{5} \text { does not give real value. }\)
\(\text { So, } \mathrm{t}^2=2+\sqrt{5}\)
Ans: b
Exp:(B) The point \(\left(\mathrm{at}^2, 2 \mathrm{bt}\right)\) lies on the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).
But \(\mathrm{t}^2=2-\sqrt{5}\) does not give real value.
So, \(\quad \mathrm{t}^2=2+\sqrt{5}\)
COMEDK-2019
Hyperbola
120688
Find the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13 .
1 \(25 \mathrm{x}^2-144 \mathrm{y}^2=900\)
2 \(144 \mathrm{x}^2-25 \mathrm{y}^2=900\)
3 \(5 \mathrm{x}^2-12 \mathrm{y}^2=30\)
4 \(12 x^2-5 y^2=30\)
Explanation:
A Let 2a and 2b be the transverse and conjugate axes and \(e\) be the eccentricity. The equation of the hyperbola is
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
We have, \(2 \mathrm{~b}=5\) and \(2 \mathrm{ae}=13\)
Now, \(b^2=a^2\left(e^2-1\right) \Rightarrow b^2=a^2 e^2-a^2\)
\(\Rightarrow \frac{25}{4}=\frac{169}{4}-\mathrm{a}^2 \Rightarrow \mathrm{a}^2=\frac{144}{4} \Rightarrow \mathrm{a}^2=36\)
Thus, the equation of the hyperbola is
\(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{25 / 4}=1 \Rightarrow 25 \mathrm{x}^2-144 \mathrm{y}^2=900\)