Hyperbola
3.0 Difference between two forms of Hyperbola
3.0 Difference between two forms of Hyperbola
Hyperbola | Conjugate Hyperbola | |
| 1. | Transverse Axis: $X$ axis$(y=0)$ Length$=2a$ | Transverse Axis: $Y$ axis$(x=0)$ Length$=2b$ |
| 2. | Conjugate Axis: $Y$ axis$(x=0)$ Length$=2b$ | Conjugate Axis: $X$ axis$(y=0)$ Length$=2a$ |
| 3. | Vertex$( \pm a,0)$ | Vertex$(0, \pm b)$ |
| 4. | Focus$( \pm ae,0)$ | Focus$(0, \pm be)$ |
| 5. | Directrix: $x = \pm \frac{a}{e}$ | Directrix: $y = \pm \frac{b}{e}$ |
| 6. | Length of Latus Rectum: $x = \pm ae$ End points: $( \pm ae, \pm \frac{{{b^2}}}{a})$ Length of latus rectum$ = \frac{{2{b^2}}}{a}$ $ = \frac{{2 \times {{(length\ of\ semi-minor\ axis)}^2}}}{{length\ of\ semi-major\ axis}}$ | Latus Rectum: $y = \pm be$ End points: $( \pm be, \pm \frac{{{a^2}}}{b})$ Length of latus rectum$ = \frac{{2{a^2}}}{b}$ $ = \frac{{2 \times {{(length\ of\ semi-minor\ axis)}^2}}}{{length\ of\ semi-major\ axis}}$ |
| 7. | Eccentricity $e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} $ | Eccentricity $e = \sqrt {1 + \frac{{{a^2}}}{{{b^2}}}} $ |
| 8. | Distance between Foci$=2ae$ | Distance between Foci$=2be$ |
| 9. |
|
|
| 10. |
|
|
| 11. |
|
|
| 12. | Transverse axis perpendicular to conjugate axis (Intersection point of transverse and conjugate axis is called centre of hyperbola) | Transverse axis perpendicular to conjugate axis (Intersection point of transverse and conjugate axis is called centre of hyperbola) |
Question 1. The hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ having eccentricity ${e_1}$ and the conjugate hyperbola $ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ having eccentricity ${e_2}$. Prove that $\frac{1}{{{e_1}^2}} + \frac{1}{{{e_2}^2}} = 1$.
Solution: As we know that, the eccentricity ${e_1}$ of hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is $\sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} $ and the eccentricity ${e_2}$ of conjugate hyperbola $ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is $\sqrt {1 + \frac{{{a^2}}}{{{b^2}}}} $.
We can write it as $${e_1}^2 = 1 + \frac{{{b^2}}}{{{a^2}}},{\text{ }}{e_2}^2 = 1 + \frac{{{a^2}}}{{{b^2}}}$$$$\frac{1}{{{e_1}^2}} + \frac{1}{{{e_2}^2}} = \frac{{{a^2}}}{{{a^2} + {b^2}}} + \frac{{{b^2}}}{{{a^2} + {b^2}}} = \frac{{{a^2} + {b^2}}}{{{a^2} + {b^2}}} = 1$$
Question 2. Find the equation of hyperbola whose foci are $(0, \pm \sqrt {10} )$ and which pass through $(2,3)$.
Solution: From the coordinates of foci given in question, we can assume that the equation of hyperbola is $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ which passes through $(2,3)$. Therefore,
$$ - \frac{4}{{{a^2}}} + \frac{9}{{{b^2}}} = 1...(1)$$
and foci is $(0, \pm be)$, $$be = \sqrt {10} $$$$e = \frac{{\sqrt {10} }}{b}$$$${e^2} = \frac{{10}}{{{b^2}}}$$$$1 + \frac{{{a^2}}}{{{b^2}}} = \frac{{10}}{{{b^2}}}$$$$10 = {a^2} + {b^2}$$$${a^2} = 10 - {b^2}...(2)$$
Solving equations $(1)$ and $(2)$, we get
$$\frac{9}{{{b^2}}} = 1 + \frac{4}{{10 - {b^2}}}$$$${b^4} - 23{b^2} + 90 = 0$$$${b^2} = \frac{{23 \pm \sqrt {169} }}{2} = \frac{{23 \pm 13}}{2} = 5{\text{ or }}18$$
From equation $(2)$, value of ${b^2}$ should be less than $10$, therefore value of ${b^2}=18$ is neglected and ${b^2}=5$. By putting the value of ${b^2}$ in equation $(2)$, we get $${a^2} = 5$$
The equation of hyperbola is $$ - \frac{{{x^2}}}{5} + \frac{{{y^2}}}{5} = 1$$
