We will learn how to solve different types of problems on hyperbola.
1. Find the position of the point (6,  5) relative to the hyperbola \(\frac{x^{2}}{9}\)  \(\frac{y^{2}}{25}\) = 1.
Solution:
The given equation is of the hyperbola is \(\frac{x^{2}}{9}\)  \(\frac{y^{2}}{25}\) = 1
We know that the point P (x\(_{1}\), y\(_{1}\)) lies outside, on or inside the hyperbola \(\frac{x^{2}}{a^{2}}\)  \(\frac{y^{2}}{b^{2}}\) = 1 according as \(\frac{x_{1}^{2}}{a^{2}}\)  \(\frac{y_{1}^{2}}{b^{2}}\)  1 < 0, = or > 0.
According to the given problem,
\(\frac{x_{1}^{2}}{a^{2}}\)  \(\frac{y_{1}^{2}}{b^{2}}\) – 1
= \(\frac{6^{2}}{9}\)  \(\frac{(5)^{2}}{25}\)  1
= \(\frac{26}{9}\)  \(\frac{25}{25}\)  1
= 4  1  1
= 2 > 0.
Therefore, the point (6,  5) lies inside the hyperbola \(\frac{x^{2}}{9}\)  \(\frac{y^{2}}{25}\) = 1
2. The coordinates of the vertices of a hyperbola are (9, 2) and (1, 2) and the distance between its two foci is 10. Find its equation and also the length of its latus rectum.
Solution:
According to the problem the ordinates of the vertices of the required hyperbola are equal. Therefore, the transverse axis of the hyperbola is parallel to axis and conjugate axis is parallel to yaxis.
The midpoint of the vertices (\(\frac{9 + 1}{2}\), \(\frac{2 + 2}{2}\)) = (5, 2)
The midpoint of the vertices is the centre of the required hyperbola.
Therefore, the centre of the required hyperbola is (5, 2)
Let the equation of the required hyperbola be \(\frac{(x  α)^{2}}{a^{2}}\)  \(\frac{(y  β)^{2}}{b^{2}}\) = 1
Now, the length its transverse axis = the distance between the two vertices i.e., the distance between the points (9, 2) and (1, 2) = 8
i.e., 2a = 8
⇒ a = 4.
Again, the distance between the two foci = 2ae = 10
⇒ ae = 5.
Now, b\(^{2}\) = a\(^{2}\)(e\(^{2}\)  1)
= a\(^{2}\)e\(^{2}\)  a\(^{2}\)
= 5\(^{2}\)  4\(^{2}\)
= 25  16
= 9
Now form the equation \(\frac{(x  α)^{2}}{a^{2}}\)  \(\frac{(y  β)^{2}}{b^{2}}\) = 1, we get,
⇒ \(\frac{(x  5)^{2}}{16}\)  \(\frac{(y  β)^{2}}{9}\) = 1
⇒ 9x\(^{2}\)  16y\(^{2}\)  90x + 64y + 17 = 0.
Therefore, the equation of the required hyperbola is
The length of the latus rectum of the hyperbola = 2 ∙ \(\frac{b^{2}}{a}\) = 2 ∙ \(\frac{9}{4}\) = \(\frac{9}{2}\) units. `More problems on hyperbola:
3. Find the equation of the hyperbola whose coordinates of the foci of a hyperbola are (± 6, 0) and its latus rectum is of 10 units.
Solution:
Let the equation of the required hyperbola be,
\(\frac{x^{2}}{a^{2}}\)  \(\frac{y^{2}}{b^{2}}\) = 1
The coordinates of the foci of the hyperbola \(\frac{x^{2}}{a^{2}}\)  \(\frac{y^{2}}{b^{2}}\) = 1 are (± ae, 0) and the lenght of its latus rectum is 2 ∙ \(\frac{b^{2}}{a}\).
According to the problem,
ae = 6 ……………….. (i) and
2 ∙ \(\frac{b^{2}}{a}\) = 10 ……………….. (ii)
Now form the above equation (ii) we get,
2b\(^{2}\) = 10a
⇒ a\(^{2}\)(e\(^{2}\)  1) = 5a , [Since, we know that, b\(^{2}\) = a\(^{2}\)(e\(^{2}\)  1)]
⇒ a\(^{2}\)e\(^{2}\)  a\(^{2}\)  5a = 0
⇒ 6\(^{2}\)  a\(^{2}\) 5a = 0 (Since, we know that, ae = 6)
⇒ a\(^{2}\) + 5a  36 = 0
⇒ a\(^{2}\) + 9a  4a  36 = 0
⇒ a(a + 9)  4(a + 9) = 0
⇒ (a + 9)(a  4) = 0
⇒ (a + 9) = 0 or (a  4) = 0
⇒ a =  9 or, a = 4
⇒ a = 9 is not possible.
Therefore, a = 4
Therefore, b\(^{2}\) = a\(^{2}\)(e\(^{2}\)  1)
= (ae)\(^{2}\)  a\(^{2}\)
= 6\(^{2}\)  4\(^{2}\)
= 36  16
= 20
Therefore, the required equation of the hyperbola is
\(\frac{x^{2}}{a^{2}}\)  \(\frac{y^{2}}{b^{2}}\) = 1
⇒ \(\frac{x^{2}}{16}\)  \(\frac{y^{2}}{20}\) = 1
⇒ (5x)\(^{2}\)  4y\(^{2}\) = 80.
`● The Hyperbola
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