Question

In each of the following find the equations of the hyperbola satisfying the given conditions:
(i) vertices (± 2, 0), foci (± 3, 0)
(ii) vertices (0, ± 5), foci (0, ± 8)
(iii) vertices (0, ± 3), foci (0, ± 5)
(iv) foci (± 5, 0), transverse axis = 8
(v) foci (0, ± 13), conjugate axis = 24
(vi) foci (± $3\sqrt{5}$, 0), the latus-rectum = 8
(vii) foci (± 4, 0), the latus-rectum = 12
(viii) vertices (± 7, 0), $e=\frac{4}{3}$
(ix) foci (0, ± $\sqrt{10}$), passing through (2, 3)
(x) foci (0, ± 12), latus-rectum = 36

Answer 1

(i) The vertices of the hyperbola are $\left(\pm 2,0\right)$ and the foci are $\left(\pm 3,0\right)$.
Thus, the value of $a=2$ and $ae=3$.
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow b^2=9-4 \\ \Rightarrow b^2=5 \end{gathered}$$
Thus, the equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{5}=1$.

(ii) The vertices of the hyperbola are $\left(0,\pm 5\right)$ and the foci are $\left(0,\pm 8\right)$
Thus, the value of $a=5$ and $ae=8$.
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow b^2=64-25 \\ \Rightarrow b^2=39 \end{gathered}$$
Thus, the equation of the hyperbola is $-\frac{x^2}{39}+\frac{y^2}{25}=1$.

(iii) The vertices of the hyperbola are $\left(0,\pm 3\right)$ and the foci are $\left(0,\pm 5\right)$.
Thus, the value of $a=3$ and $ae=5$.
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow b^2=25-9 \\ \Rightarrow b^2=16 \end{gathered}$$
Thus, the equation of the hyperbola is $-\frac{x^2}{16}+\frac{y^2}{9}=1$.

(iv) The foci of the hyperbola are $\left(\pm 5,0\right)$ and the transverse axis is 8.
Thus, the value of $ae=5$ and 2a = 8.
$\Rightarrow a=4$
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow b^2=25-16 \\ \Rightarrow b^2=9 \end{gathered}$$
Thus, the equation of the hyperbola is$\frac{x^2}{16}-\frac{y^2}{9}=1$.

(v) The foci of the hyperbola are $\left(0,\pm 13\right)$ and the conjugate axis is 24.
Thus, the value of $ae=13$ and 2b = 24.
⇒ b = 12

Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow a^2=169-144 \\ \Rightarrow a^2=25 \end{gathered}$$
Thus, the equation of the hyperbola is$-\frac{x^2}{144}+\frac{y^2}{25}=1$.

(vi) The foci of the hyperbola are $\left(\pm 3\sqrt{5},0\right)$ and the latus rectum is 8.
Thus, the value of $ae=3\sqrt{5}$
$$\begin{gathered} \mathrm{and}\frac{2b^2}{a}=8 \\ \Rightarrow b^2=4a \end{gathered}$$
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow 4a=45-a^2 \\ \Rightarrow a^2+4a-45=0 \\ \Rightarrow \left(a-5\right)\left(a+9\right)=0 \\ \Rightarrow a=-9,5 \end{gathered}$$
$b^2=-36\mathrm{or}20$
Since negative value is not possible, it is equal to 20.
Thus, the equation of the hyperbola is$\frac{x^2}{25}-\frac{y^2}{20}=1$.

(vii) The foci of the hyperbola are $\left(\pm 4,0\right)$ and the latus rectum is 12.
Thus, the value of $ae=4$
$$\begin{gathered} \mathrm{and}\frac{2b^2}{a}=12 \\ \Rightarrow b^2=6a \end{gathered}$$
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow 6a=16-a^2 \\ \Rightarrow a^2+6a-16=0 \\ \Rightarrow \left(a-2\right)\left(a+8\right)=0 \\ \Rightarrow a=2,\mathrm{or}-8 \end{gathered}$$
$b^2=12\mathrm{or}-48$
Since negative value is not possible, its value is 12.
Thus, the equation of the hyperbola is$\frac{x^2}{4}-\frac{y^2}{12}=1$.

(viii) The vertices of hyperbola are $\left(\pm 7,0\right)$ and eccentricity is $\frac{4}{3}$
Thus, the value of $a=7$.
Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow b^2=49\left(\frac{16}{9}-1\right) \\ \Rightarrow b^2=49\times \frac{7}{9}=\frac{343}{9} \end{gathered}$$
Thus, the equation of the hyperbola is$\frac{x^2}{49}-\frac{9y^2}{343}=1$.

(ix) The foci of hyperbola are $\left(0,\pm \sqrt{10}\right)$ that pass through $\left(2,3\right)$.

$$\begin{gathered} \mathrm{Thus,\; the}\mathrm{value}ofae=\sqrt{10}. \\ \mathrm{By}\mathrm{squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ {\left(ae\right)}^2=10 \\ \Rightarrow a^2+b^2=10 \\ \Rightarrow b^2=10-a^2 \end{gathered}$$
Let the equation of the hyperbola be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$.
It passes through $\left(2,3\right)$.
$$\begin{gathered} \Rightarrow \frac{3^2}{a^2}-\frac{2^2}{10-a^2}=1 \\ \Rightarrow 90-9a^2-4a^2=10a^2-a^4 \\ \Rightarrow a^4-23a^2+90=0 \\ \Rightarrow \left(a^2-18\right)\left(a^2-5\right)=0 \\ \Rightarrow a^2=18,5 \end{gathered}$$
Now, $b^2=-8\mathrm{or}5$
If we neglect the negative value, then b² = 5.
Thus, the equation of the hyperbola is$\frac{y^2}{5}-\frac{x^2}{5}=1$.

(x) The foci of the hyperbola are $\left(0,\pm 12\right)$ and the latus rectum is 36.
Thus, the value of $ae=12$.
$$\begin{gathered} \mathrm{and}\frac{2b^2}{a}=36 \\ \Rightarrow b^2=18a \end{gathered}$$

Now, using the relation $b^2=a^2(e^2-1)$, we get:
$$\begin{gathered} \Rightarrow 18a=144-a^2 \\ \Rightarrow a^2+18a-144=0 \\ \Rightarrow \left(a+24\right)\left(a-6\right)=0 \\ \Rightarrow a=-24,\mathrm{or}6 \end{gathered}$$
$b^2=-432\mathrm{or}108(\mathrm{but}\mathrm{negative}\mathrm{value}\mathrm{is}\mathrm{not}\mathrm{possible})$

Thus, the equation of the hyperbola is$\frac{y^2}{36}-\frac{x^2}{108}=1$.