Q1. \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\)

Solution

The given equation of the hyperbola is \[ \begin{aligned} \dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1 \end{aligned} \] Comparing this with the standard form \(\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\), we identify \[ \begin{aligned} a^{2}&=16 \\ a&=4 \\ b^{2}&=9 \\ b&=3 \end{aligned} \]

For a hyperbola of this form, the relation between \(a\), \(b\), and \(c\) is \(c^{2}=a^{2}+b^{2}\). Hence, \[ \begin{aligned} c&=\sqrt{a^{2}+b^{2}} \\ &=\sqrt{16+9} \\ &=5 \end{aligned} \]

The foci of the hyperbola lie on the transverse axis and are given by \((\pm c,0)\). Therefore, \[ \begin{aligned} &(\pm c,0) \\ &=(\pm 5,0) \end{aligned} \]

The vertices of the hyperbola are the endpoints of the transverse axis and are given by \((\pm a,0)\). Hence, \[ \begin{aligned} &(\pm a,0) \\ &=(\pm 4,0) \end{aligned} \]

The eccentricity of a hyperbola is defined as the ratio of the distance of a focus from the centre to the semi-transverse axis, that is \(e=\dfrac{c}{a}\). Thus, \[ \begin{aligned} e&=\dfrac{c}{a}\\ &=\dfrac{5}{4} \end{aligned} \]

The length of the latus rectum of a hyperbola \(\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\) is given by \(\dfrac{2b^{2}}{a}\). Substituting the values of \(a\) and \(b\), \[ \begin{aligned} \dfrac{2b^{2}}{a} &=\dfrac{2\times 9}{4} \\ &=\dfrac{9}{2} \end{aligned} \]

Q2. \(\dfrac{y^2}{9}-\dfrac{x^2}{27}=1\)

Solution

The given equation of the hyperbola is \[ \begin{aligned} \dfrac{y^{2}}{9}-\dfrac{x^{2}}{27}=1 \end{aligned} \] This is compared with the standard form \(\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1\), which represents a hyperbola with transverse axis along the \(y\)-axis. From this comparison, we obtain \[ \begin{aligned} a^{2}=9 \\ a=3 \\ b^{2}=27 \\ b=3\sqrt{3} \end{aligned} \]

For a hyperbola, the parameters \(a\), \(b\), and \(c\) are related by \(c^{2}=a^{2}+b^{2}\). Substituting the known values, \[ \begin{aligned} c=\sqrt{a^{2}+b^{2}} \\ =\sqrt{9+27} \\ =\sqrt{36} \\ =6 \end{aligned} \]

Since the transverse axis is along the \(y\)-axis, the foci lie on the \(y\)-axis and their coordinates are given by \((0,\pm c)\). Hence, \[ \begin{aligned} (0,\pm c) \\ =(0,\pm 6) \end{aligned} \]

The vertices of the hyperbola are the endpoints of the transverse axis and are given by \((0,\pm a)\). Therefore, \[ \begin{aligned} (0,\pm a) \\ =(0,\pm 3) \end{aligned} \]

The eccentricity of a hyperbola is defined as the ratio of the distance of a focus from the centre to the semi-transverse axis, that is \(e=\dfrac{c}{a}\). Thus, \[ \begin{aligned} e=\dfrac{c}{a} \\ =\dfrac{6}{3} \\ =2 \end{aligned} \]

The length of the latus rectum of the hyperbola \(\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1\) is given by \(\dfrac{2b^{2}}{a}\). Substituting the values of \(a\) and \(b\), \[ \begin{aligned} \dfrac{2b^{2}}{a} \\ =\dfrac{2\times 27}{3} \\ =18 \end{aligned} \]

Q3. \(9y^2 – 4x^2 = 36\)

Solution

9y² – 4x² = 36

Divide both sides by 36 to bring the equation to standard form.

$$\begin{aligned} \frac{9y^{2}}{36}-\frac{4x^{2}}{36}=1\\ \frac{y^{2}}{4}-\frac{x^{2}}{9}=1 \end{aligned}$$

This is the standard form of a hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), which opens vertically along the y-axis. Here, the transverse axis is vertical.

Thus, \(a^2 = 4\) so \(a = 2\), and \(b^2 = 9\) so \(b = 3\).

The distance to each focus from the center is \(c = \sqrt{a^2 + b^2}\).

$$\begin{aligned} c&=\sqrt{4+9}\\ &=\sqrt{13} \end{aligned}$$ Foci

The foci are located at \((0, \pm c)\) since the transverse axis is vertical.

$$\left(0, \pm \sqrt{13}\right)$$ Vertices

The vertices are at \((0, \pm a)\), the endpoints of the transverse axis.

$$\left(0, \pm 2\right)$$ Eccentricity

The eccentricity \(e\) measures how much the hyperbola deviates from a circle, given by \(e = \frac{c}{a}\) for hyperbolas.

$$\begin{aligned} e&=\frac{\sqrt{13}}{2} \end{aligned}$$ Latus Rectum

The length of the latus rectum, a chord through the focus perpendicular to the transverse axis, is \(\frac{2b^2}{a}\).

$$\begin{aligned} \frac{2b^{2}}{a}&=\frac{2 \times 9}{2}\\ &=9 \end{aligned}$$

Q4. \(16x^2 – 9y^2 = 576\)

Solution

16x² – 9y² = 576

Divide both sides by 576 to obtain the standard form of the hyperbola.

$$\begin{aligned} \frac{16x^{2}}{576}-\frac{9y^{2}}{576}=1\\ \frac{x^{2}}{36}-\frac{y^{2}}{64}=1 \end{aligned}$$

This matches the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), indicating a horizontal transverse axis along the x-axis.

Therefore, \(a^2 = 36\) so \(a = 6\), and \(b^2 = 64\) so \(b = 8\).

The distance from the center to each focus is \(c = \sqrt{a^2 + b^2}\).

$$\begin{aligned} c&=\sqrt{36+64}\\ &=\sqrt{100}\\ &=10 \end{aligned}$$

Foci of Hyperbola

The foci lie on the transverse axis at \(( \pm c, 0 )\) for a horizontally oriented hyperbola.

$$\left( \pm 10, 0 \right)$$

Vertices of Hyperbola

The vertices are the endpoints of the transverse axis at \(( \pm a, 0 )\).

$$\left( \pm 6, 0 \right)$$

Eccentricity of Hyperbola

Eccentricity \(e = \frac{c}{a}\) quantifies the shape's elongation beyond an ellipse.

$$\begin{aligned} e&=\frac{10}{6}\\ &=\frac{5}{3} \end{aligned}$$

Latus Rectum

The latus rectum length is \(\frac{2b^2}{a}\), passing through the focus perpendicular to the transverse axis.

$$\begin{aligned} \frac{2b^{2}}{a}&=\frac{2 \times 64}{6}\\ &=\frac{128}{6}\\ &=\frac{64}{3} \end{aligned}$$

Q5. \(5y^2 – 9x^2 = 36\)

Solution

5y² – 9x² = 36

Divide throughout by 36 to convert into standard form.

$$\begin{aligned} \frac{5y^{2}}{36}-\frac{9x^{2}}{36}=1\\ \frac{y^{2}}{36/5}-\frac{x^{2}}{4}=1 \end{aligned}$$

The equation is in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), a hyperbola opening upwards and downwards with vertical transverse axis.

Hence, \(a^2 = \frac{36}{5}\) so \(a = \frac{6}{\sqrt{5}}\), and \(b^2 = 4\) so \(b = 2\).

Compute \(c = \sqrt{a^2 + b^2}\) for focus locations.

$$\begin{aligned} c&=\sqrt{\frac{36}{5} + 4}\\ &=\sqrt{\frac{36}{5} + \frac{20}{5}}\\ &=\sqrt{\frac{56}{5}}\\ &=\frac{\sqrt{56}}{\sqrt{5}}\\ &=\frac{2\sqrt{14}}{\sqrt{5}} \end{aligned}$$

Foci of Hyperbola

For vertical transverse axis, foci are at \((0, \pm c)\).

$$\left( 0, \pm \frac{2\sqrt{14}}{\sqrt{5}} \right)$$

Vertices of Hyperbola

Vertices mark the ends of the transverse axis at \((0, \pm a)\).

$$\left( 0, \pm \frac{6}{\sqrt{5}} \right)$$

Eccentricity

Eccentricity \(e = \frac{c}{a}\) is always greater than 1 for hyperbolas.

$$\begin{aligned} e&=\frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}}\\ &=\frac{2\sqrt{14}}{6}\\ &=\frac{\sqrt{14}}{3} \end{aligned}$$

Latus Rectum

Latus rectum length \(\frac{2b^2}{a}\) passes through focus perpendicular to transverse axis.

$$\begin{aligned} \frac{2b^{2}}{a}&=\frac{2 \times 4}{\frac{6}{\sqrt{5}}}\\ &=\frac{8 \times \sqrt{5}}{6}\\ &=\frac{4\sqrt{5}}{3} \end{aligned}$$

Q6. \( 49y^2 – 16x^2 = 784\)

Solution

49y² – 16x² = 784

Divide both sides by 784 to get the standard form.

$$\begin{aligned} \frac{49y^{2}}{784}-\frac{16x^{2}}{784}=1\\ \frac{y^{2}}{16}-\frac{x^{2}}{49}=1 \end{aligned}$$

This is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), a hyperbola with vertical transverse axis opening up and down.

So, \(a^2 = 16\) gives \(a = 4\), and \(b^2 = 49\) gives \(b = 7\).

Find \(c = \sqrt{a^2 + b^2}\) to locate the foci.

$$\begin{aligned} c&=\sqrt{16 + 49}\\ &=\sqrt{65} \end{aligned}$$

Foci of Hyperbola

Foci are positioned at \((0, \pm c)\) along the vertical transverse axis.

$$\left( 0, \pm \sqrt{65} \right)$$

Vertices of Hyperbola

Vertices lie at the ends of the transverse axis, \((0, \pm a)\).

$$\left( 0, \pm 4 \right)$$

Eccentricity

The eccentricity \(e = \frac{c}{a}\) describes the hyperbola's openness.

$$\begin{aligned} e&=\frac{\sqrt{65}}{4} \end{aligned}$$

Latus Rectum

Length of latus rectum is \(\frac{2b^2}{a}\), the chord through focus perpendicular to transverse axis.

$$\begin{aligned} \frac{2b^{2}}{a}&=\frac{2 \times 49}{4}\\ &=\frac{98}{4}\\ &=24.5 \end{aligned}$$

Q7. Vertices (± 2, 0), foci (± 3, 0)

Solution

Vertices (±2, 0), foci (±3, 0)

Given vertices at (±2, 0), the hyperbola has horizontal transverse axis centered at origin, so \(a = 2\).

Foci at (±3, 0) give \(c = 3\).

For hyperbolas, the relationship \(c^2 = a^2 + b^2\) holds, where \(b^2\) determines the asymptotes.

$$\begin{aligned} b^2&=c^2 - a^2\\ &=3^2 - 2^2\\ &=9 - 4\\ &=5 \end{aligned}$$

Equation of Hyperbola

The standard form for horizontal transverse axis is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

$$\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$$

Q8. Vertices (0, ± 5), foci (0, ± 8)

Solution

Vertices (0, ±5), foci (0, ±8)

Vertices at (0, ±5) indicate vertical transverse axis centered at origin, so \(a = 5\).

Foci at (0, ±8) give \(c = 8\).

Hyperbolas satisfy \(c^2 = a^2 + b^2\), where \(b^2\) defines conjugate axis length.

$$\begin{aligned} b^2&=c^2 - a^2\\ &=8^2 - 5^2\\ &=64 - 25\\ &=39 \end{aligned}$$

Equation of Hyperbola

Standard form for vertical transverse axis is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

$$\frac{y^{2}}{25} - \frac{x^{2}}{39} = 1$$

Q9. Vertices (0, ± 3), foci (0, ± 5)

Solution

Vertices (0, ±3), foci (0, ±5)

Vertices at (0, ±3) show vertical transverse axis centered at origin, so \(a = 3\).

Foci located at (0, ±5) give \(c = 5\).

The hyperbola relation \(c^2 = a^2 + b^2\) allows finding \(b^2\).

$$\begin{aligned} b^2&=c^2 - a^2\\ &=5^2 - 3^2\\ &=25 - 9\\ &=16 \end{aligned}$$

Equation of Hyperbola

Vertical transverse axis uses standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

$$\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1$$

Q10. Foci (± 5, 0), the transverse axis is of length 8.

Solution

Foci (±5, 0), transverse axis length 8

Transverse axis length of 8 means \(2a = 8\), so \(a = 4\).

Foci at (±5, 0) indicate horizontal transverse axis with \(c = 5\).

Using the hyperbola property \(c^2 = a^2 + b^2\) to find \(b^2\).

$$\begin{aligned} b^2&=c^2 - a^2\\ &=5^2 - 4^2\\ &=25 - 16\\ &=9 \end{aligned}$$

Equation of Hyperbola

Horizontal transverse axis follows standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

$$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$$

Q11. Foci (0, ±13), the conjugate axis is of length 24.

Solution

Foci (0, ±13), conjugate axis length 24

Conjugate axis length is \(2b = 24\), so \(b = 12\) and \(b^2 = 144\).

Foci at (0, ±13) confirm vertical transverse axis with \(c = 13\).

Hyperbola definition gives \(c^2 = a^2 + b^2\), solve for \(a^2\).

$$\begin{aligned} a^2&=c^2 - b^2\\ &=13^2 - 12^2\\ &=169 - 144\\ &=25 \end{aligned}$$

Thus, \(a = 5\) since \(a > 0\).

Equation of Hyperbola

Vertical transverse axis uses form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

$$\frac{y^{2}}{25} - \frac{x^{2}}{144} = 1$$

Q12. Foci \((± 3\sqrt{5}, 0)\), the latus rectum is of length 8.

Solution

Foci \(( \pm 3\sqrt{5}, 0 )\), latus rectum length 8

Latus rectum length \(\frac{2b^2}{a} = 8\), so \(b^2 = 4a\).

Foci at \((\pm 3\sqrt{5}, 0)\) indicate horizontal transverse axis with \(c = 3\sqrt{5}\).

Substitute into \(c^2 = a^2 + b^2\) and solve the quadratic.

$$\begin{aligned} c^2&=a^2 + b^2\\ (3\sqrt{5})^2&=a^2 + 4a\\ 45&=a^2 + 4a\\ a^2 + 4a - 45&=0\\ a&=\frac{-4 \pm \sqrt{16 + 180}}{2}\\ &=\frac{-4 \pm \sqrt{196}}{2}\\ &=\frac{-4 \pm 14}{2} \end{aligned}$$

Positive root \(a = 5\) (discard negative), then \(b^2 = 4 \times 5 = 20\).

Equation of Hyperbola

Horizontal transverse axis gives standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

$$\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$$

Q13. Foci (± 4, 0), the latus rectum is of length 12

Solution

Foci (±4, 0), latus rectum length 12

Latus rectum \(\frac{2b^2}{a} = 12\) implies \(b^2 = 6a\).

Foci (±4, 0) confirm horizontal transverse axis, \(c = 4\).

Apply \(c^2 = a^2 + b^2\) and solve resulting quadratic equation.

$$\begin{aligned} c^2&=a^2 + b^2\\ 16&=a^2 + 6a\\ a^2 + 6a - 16&=0 \end{aligned}$$ $$\begin{aligned} a&=\frac{-6 \pm \sqrt{36 + 64}}{2}\\ &=\frac{-6 \pm 10}{2} \end{aligned}$$

Take positive value \(a = 2\), then \(b^2 = 6 \times 2 = 12\).

Equation of Hyperbola

Horizontal transverse axis yields form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

$$\frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$$

Q14. vertices \((± 7,0),\; e =\dfrac{4}{3}\)

Solution

Vertices (±7, 0), \(e = \frac{4}{3}\)

Vertices (±7, 0) give horizontal transverse axis, \(a = 7\).

Eccentricity \(e = \frac{c}{a} = \frac{4}{3}\), so compute \(c\).

$$\begin{aligned} c&=e \cdot a\\ &=\frac{4}{3} \times 7\\ &=\frac{28}{3} \end{aligned}$$

From \(c^2 = a^2 + b^2\), find \(b^2\).

$$\begin{aligned} b^2&=c^2 - a^2\\ &=\left( \frac{28}{3} \right)^2 - 7^2\\ &=\frac{784}{9} - \frac{441}{9}\\ &=\frac{343}{9} \end{aligned}$$

Equation of Hyperbola

Standard form for horizontal transverse axis is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

$$\frac{x^{2}}{49} - \frac{y^{2}}{\frac{343}{9}} = 1$$

Or equivalently, \(\frac{x^2}{49} - \frac{9y^2}{343} = 1\).

Q15. Foci \((0, ±\sqrt{10})\), passing through (2,3)

Solution

Hyperbola passing through (2,3), foci \((0, \pm \sqrt{10})\)

Foci indicate vertical transverse axis, \(c = \sqrt{10}\).

Hyperbola relation \(c^2 = a^2 + b^2\) gives \(10 = a^2 + b^2\), so \(a^2 = 10 - b^2\).

Point (2,3) satisfies \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

$$\begin{aligned} \frac{3^2}{a^2} - \frac{2^2}{b^2} &= 1\\ \frac{9}{a^2} - \frac{4}{b^2} &= 1 \end{aligned}$$ $$\begin{aligned} 9b^2 - 4a^2 &= a^2 b^2\\ 9b^2 - 4(10 - b^2) &= (10 - b^2) b^2\\ 9b^2 - 40 + 4b^2 &= 10b^2 - b^4\\ 13b^2 - 40 &= 10b^2 - b^4 \end{aligned}$$ $$\begin{aligned} b^4 + 3b^2 - 40 &= 0 \end{aligned}$$

Let \(x = b^2\), solve \(x^2 + 3x - 40 = 0\).

$$\begin{aligned} x^2 + 8x - 5x - 40 &= 0\\ x(x + 8) - 5(x + 8) &= 0\\ (x + 8)(x - 5) &= 0 \end{aligned}$$

Positive root \(x = 5\), so \(b^2 = 5\), then \(a^2 = 10 - 5 = 5\).

Equation of Hyperbola

$$\frac{y^{2}}{5} - \frac{x^{2}}{5} = 1$$