Introduction to hyperbola - class-XI

For the rectangular hyperbola xy = c^2, the standard form is X^2 - Y^2 = 2c^2 (after rotation). The length of the latus rectum for this hyperbola is 2*sqrt(2)*c.

The area of the quadrilateral formed by tangents at the ends of the latus rectum is 2b^2. Setting this equal to (ae)^2 and using b^2 = a^2(e^2 - 1) leads to e^3 = 2sqrt(2).

Standard equation of the hyperbola is $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}$

The equation is of the form (distance from focus)^2 = e^2 * (distance from directrix)^2. By identifying the focus and directrix, one can determine the latus rectum equation.

The general equation of an ellipse is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and that of a hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$
Using this, we get that the above equation is an ellipse if $10 - \lambda > 0$ and $6 - \lambda > 0$. The combined solution gives $\lambda < 6.$
For a hyperbola, the coefficient of $x^2$ and $y^2$ must be of opposite sign. Hence, 
 $10 - \lambda > 0$ and $6 - \lambda < 0$ which gives $6 < \lambda < 10$

The given differential equation can be rearranged into a form that describes a family of curves, which are hyperbolas or parabolas depending on the integration constants.

$Standard, hyperbola, : $

The focus of the ellipse x^2/25 + y^2/16 = 1 is (3, 0). If the hyperbola passes through (3, 0) and shares axes, its equation is x^2/a^2 - y^2/b^2 = 1. Using the eccentricity product condition, one can solve for the focus.

For the hyperbola xy = 4, the axes are rotated by 45 degrees. The foci are located at (2sqrt(2), 2sqrt(2)) and (-2sqrt(2), -2sqrt(2)).

The eccentricity e = sqrt(1 + tan^2(theta)) = sec(theta). If e > 2, then sec(theta) > 2, so cos(theta) < 1/2. The latus rectum length is 2*sin^2(theta)/cos(theta). Solving this interval gives (3, infinity).

We have,

The locus is nothing but hyperbola.
Difference of distance of a point from foci $=2a$ 

For a parabola the eccentricity is $1$

The parametric equations x = asec(t) and y = btan(t) define a hyperbola. The given equations are a variation of this, leading to an eccentricity of sqrt(2).

Given expression,$\displaystyle 3{ x }^{ 2 }-2xy+{ y }^{ 2 }=0$ 
As Coefficient of $\displaystyle xy$ is not zero,It will not be a circle and hyperbola.
Let $\displaystyle \frac { y }{ x } =m$

It is given that
$log(\theta-\theta _{0})=kt $ where k is a constant of proportionality.
Hence
$e^{kt}=\theta-\theta _{0}$
Or
$\triangle \theta=e^{kt}$
The graph of $\triangle\theta$ vs $t$ will therefore be a hyperbola.

Given equation will represent hyperbola if
$h^2> ab$
$\Rightarrow \lambda^2, >, (\lambda, +, 2), (\lambda, -, 1)$
$\Rightarrow, \lambda, <, 2$
Also $\Delta, \neq, 0$
$\Rightarrow, -2(\lambda^2, +, \lambda, -, 2), -, 4(\lambda, -, 1), +, 2 \lambda^2, \neq, 0$
$\Rightarrow, \lambda, \neq, \displaystyle \frac{4}{3}$.
Hence option 'B' is correct.

Let the equation of hyperbola be $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$

Clearly, $|SP-S'P|=2a=12$
Thus statement 1 is correct.
Also statement 2 is correct and followed by statement 1.

The equation of hyperbola is $\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

The Foci of hyperbola are $(\pm8,0)$, hence Foci lie on $x$ - axis. 

$ e^2 = \dfrac {a^2 + b^2}{a^2} $ and $ e'^{2} = \dfrac {a^2 + b^2}{b^2} $

$ \Rightarrow \dfrac {1}{e^2} + \dfrac {1}{e'^2} = \dfrac {a^2}{a^2+b^2} + \dfrac {b^2}{a^2 + b^2} = 1 $

Let the centre of hyperbola be $(\alpha , \beta)$

Let $(h,k)$ be the point of intersection of the given lines. Then,

AFC curve is downward sloping because fixed costs are distributed over large volume when quantity produced increases.

Consider equation of a hyperbola as $F=ax^2+2by+cy^2+2dx+2ey+f=0$

Given that the length of the latus rectum is $8$ and length of the conjugate axis is equal to half the distance between the foci.

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be a point on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 }
}{ { b }^{ 2 } } =1$. Then,
$\cfrac { { { x } _{ 1 } }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1$
The chord of contact of tangents from $P$ to the hyperbola $\cfrac { { x

}^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =2$ is
$\cfrac

{ { x }{ x } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { y }{ y } _{ 1 } }{ { b

}^{ 2 } } =2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

From given expression, $\displaystyle { x }^{ 2 }-5xy+6{ y }^{ 2 }=0$ are pair of straight lines $\displaystyle y=\frac { x }{ 2 } $ and $ y=\dfrac { x }{ 3 } $
Now, $\displaystyle a{ x }^{ 2 }+b{ y }^{ 2 }+c=0$ will be cirlce of radius $\displaystyle \sqrt { -\frac { c }{ a }  } $.

Given ellipse is, $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$
Eccentricity of the ellipse is, $\displaystyle e _e  =\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}$
So the foci of the ellipse is, $(\pm ae _e,0)=(\pm 3 , 0)$
Let eccentricity of the required  hyperbola is $e _h$ and semi major and minor axes are $a$ and $b$, so the equation of hyperbola is, $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given hyperbola passes trough $(\pm 3,0)\Rightarrow \displaystyle \frac{9}{a^2}-\frac{0}{b^2}=1\Rightarrow a^2 = 9$
Also given that $\displaystyle  e _e\times e _h = 1$ $\Rightarrow e _h=\cfrac{5}{3}$ $\Rightarrow$ $b^2 =a^2(e _h^2-1)=16 $
Hence required hyperbola is, $\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1$
And foci of the hyperbola is, $(\pm 5,0)$

Given, ${x}^{2}+9=2{y}^{2}$