Question

Find the equation of the hyperbola whose
(i) foci are (6, 4) and (−4, 4) and eccentricity is 2.
(ii) vertices are (−8, −1) and (16, −1) and focus is (17, −1)
(iii) foci are (4, 2) and (8, 2) and eccentricity is 2.
(iv) vertices are at (0 ± 7) and foci at $\left(0,\pm \frac{28}{3}\right)$.
(v) vertices are at (± 6, 0) and one of the directrices is x = 4. [NCERT EXEMPLAR]
(vi) foci at (± 2, 0) and eccentricity is 3/2. [NCERT EXEMPLAR]

Answer 1

(i) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are $\left(\frac{6-4}{2},\frac{4+4}{2}\right),\mathrm{i}.\mathrm{e}.\left(1,4\right)$.
Let 2a and 2b be the length of the transverse and conjugate axis and let e be the eccentricity.

$\Rightarrow \frac{{\left(x-1\right)}^2}{a^2}-\frac{{\left(y-4\right)}^2}{b^2}=1$
Distance between the two focii = 2ae

$$\begin{gathered} 2ae=\sqrt{{\left(6+4\right)}^2+{\left(4-4\right)}^2} \\ \Rightarrow 2ae=10 \\ \Rightarrow ae=5 \\ \Rightarrow a=\frac{5}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Also,}{b}^2={\left(ae\right)}^2-{\left(a\right)}^2 \\ \Rightarrow b^2=25-\left(\frac{25}{4}\right) \\ \Rightarrow b^2=\frac{75}{4} \end{gathered}$$

Equation of the hyperbola is given below:
$$\begin{gathered} \frac{4{\left(x-1\right)}^2}{25}-\frac{4{\left(y-4\right)}^2}{75}=1 \\ \Rightarrow \frac{4\left(x^2-2x+1\right)}{25}-\frac{\left(y^2-8y+16\right)}{75}=1 \\ \Rightarrow \frac{\left(4x^2-8x+4\right)}{25}-\frac{\left(4y^2-32y+64\right)}{75}=1 \\ \Rightarrow 3\left(4x^2-8x+4\right)-\left(4y^2-32y+64\right)=75 \\ \Rightarrow 12x^2-24x+12-4y^2+32y-64=75 \\ \Rightarrow 12x^2-24x-4y^2+32y-127=0 \end{gathered}$$

(ii) The centre of the hyperbola is given below:
$\left(\frac{-8+16}{2},\frac{-1-1}{2}\right)=\left(4,-1\right)$
If the other focus is $S\text{'}\left(m,n\right)$, then it is calculated in the following way:
$$\begin{gathered} 4=\frac{17+m}{2} \\ \Rightarrow m=-9 \\ And-1=\frac{-1+n}{2} \\ \Rightarrow n=-1 \end{gathered}$$
Thus, the other focus is $\left(-9,-1\right)$.

$$\begin{gathered} \mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{vertices}: \\ 2a=\sqrt{{\left(16+8\right)}^2+{\left(-1+1\right)}^2} \\ \Rightarrow 2a=24 \\ \Rightarrow a=12 \end{gathered}$$
Distance between the foci:
$$\begin{gathered} 2c=\sqrt{{\left(17+9\right)}^2+{\left(-1+1\right)}^2} \\ \Rightarrow 2c=26 \\ \Rightarrow c=13 \end{gathered}$$

$$\begin{gathered} \mathrm{Also,}{c}^2=a^2+b^2 \\ \Rightarrow b^2=169-144 \\ \Rightarrow b^2=25 \end{gathered}$$

Equation of the hyperbola is given below:
$$\begin{gathered} \frac{{\left(x-4\right)}^2}{{\displaystyle 144}}-\frac{{\left(y+1\right)}^2}{{\displaystyle 25}}=1 \\ \Rightarrow \frac{x^2-8x+16}{{\displaystyle 144}}-\frac{\left(y^2+2y+1\right)}{{\displaystyle 25}}=1 \\ \Rightarrow 25x^2-200x+400-\left(144y^2+288y+144\right)=3600 \\ \Rightarrow 25x^2-200x-144y^2-288y-3344=0 \end{gathered}$$

(iii) The centre of the hyperbola is the midpoint of the line joining the two focii.
So, the coordinates of the centre are $\left(\frac{4+8}{2},\frac{2+2}{2}\right),\mathrm{i}.\mathrm{e}.\left(6,2\right)$.
Let 2a and 2b be the length of the transverse and conjugate axis. Let e be the eccentricity.

$\Rightarrow \frac{{\left(x-6\right)}^2}{a^2}-\frac{{\left(y-2\right)}^2}{b^2}=1$
Distance between the two focii = 2ae

$$\begin{gathered} 2ae=\sqrt{{\left(4-8\right)}^2+{\left(2-2\right)}^2} \\ \Rightarrow 2ae=4 \\ \Rightarrow ae=2 \\ \Rightarrow a=1 \end{gathered}$$

$$\begin{gathered} \mathrm{Also,}{b}^2={\left(ae\right)}^2-{\left(a\right)}^2 \\ \Rightarrow b^2=4-1 \\ \Rightarrow b^2=3 \end{gathered}$$

Equation of the hyperbola:
$$\begin{gathered} \frac{{\left(x-6\right)}^2}{1}-\frac{{\left(y-2\right)}^2}{3}=1 \\ \Rightarrow \frac{\left(x^2-12x+36\right)}{1}-\frac{\left(y^2-4y+4\right)}{3}=1 \\ \Rightarrow 3\left(x^2-12x+36\right)-\left(y^2-4y+4\right)=3 \\ \Rightarrow 3x^2-36x+108-y^2+4y-4=3 \\ \Rightarrow 3x^2-y^2-36x+4y+101=0 \end{gathered}$$


(iv) The Vertices of the hyperbola are $\left(0,\pm 7\right)$.
∴ $b=7$
The foci is $\left(0,\pm \frac{28}{3}\right)$.
∴ $be=\frac{28}{3}$

$$\begin{gathered} \mathrm{Also,}{a}^2=b^2\left(e^2-1\right) \\ \Rightarrow a^2={\left(\frac{28}{3}\right)}^2-49 \\ \Rightarrow a^2=\frac{343}{9} \end{gathered}$$
Therefore, the equation of the hyperbola is$-\frac{9x^2}{{\displaystyle 343}}+\frac{y^2}{{\displaystyle 49}}=1$.

(v) The Vertices of the hyperbola are $\left(\pm 6,0\right)$.
∴ $a=6$
⇒ a² = 36
Now, x = 4
$$\begin{gathered} \frac{a}{e}=4 \\ \Rightarrow e=\frac{3}{2}\left[\because a=6\right] \end{gathered}$$
Now,
$$\begin{gathered} {\left(ae\right)}^2=a^2+b^2 \\ \Rightarrow {\left(6\times \frac{3}{2}\right)}^2=6^2+b^2 \\ \Rightarrow 81-36=b^2 \\ \Rightarrow b^2=45 \end{gathered}$$

Therefore, the equation of the hyperbola is$\frac{x^2}{{\displaystyle 36}}-\frac{y^2}{{\displaystyle 45}}=1$.

(vi) The foci of the hyperbola are $\left(\pm 2,0\right)$.
∴ $$\begin{gathered} ae=2 \\ \Rightarrow a=2\times \frac{2}{3}=\frac{4}{3} \\ \Rightarrow a^2=\frac{16}{9} \end{gathered}$$
Now,
$$\begin{gathered} {\left(ae\right)}^2=a^2+b^2 \\ \Rightarrow {\left(2\right)}^2={\left(\frac{4}{3}\right)}^2+b^2 \\ \Rightarrow 4-\frac{16}{9}=b^2 \\ \Rightarrow b^2=\frac{20}{9} \end{gathered}$$

Therefore, the equation of the hyperbola is given by
$$\begin{gathered} \frac{9x^2}{{\displaystyle 16}}-\frac{9y^2}{{\displaystyle 20}}=1 \\ \Rightarrow \frac{x^2}{{\displaystyle 4}}-\frac{y^2}{{\displaystyle 5}}=\frac{4}{9} \end{gathered}$$