Tag: parametric equation of the hyperbola

Questions Related to parametric equation of the hyperbola

If foci of $\dfrac {x^2}{a^2}-\dfrac {y^2}{b^2}=1$ coincide with the foci of $\dfrac {x^2}{25}+\dfrac {y^2}{9}=1$ and eccentricity of the hyperbola is 2, then :

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $a^2+b^2=16$

  2. there is no director circle to the hyperbola

  3. centre of the director circle is $(0, 0)$

  4. length of latus rectum of the hyperbola $=12$

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Correct Option: A,B,D
Explanation:

$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1$
which is an ellipse.
For ellipse, $a=5$ , $b=3$ 
$\Rightarrow \displaystyle e=\sqrt {\frac {25-9}{25}}=\frac {4}{5}$
$\therefore ae=4$
Hence, the foci are $(-4, 0)$ and $(4, 0)$.
For the hyperbola,
$ae=4, e=2$
$\Rightarrow a=2$
$b^2=4(4-1)=12$
$\Rightarrow b=\sqrt {12}$
$\Rightarrow a^2+b^2=16$
Since $b^2>a^2$, so there is no director circle to the hyperbola.
Length of latus rectum $\displaystyle=\frac{2b^2}{a}=12$

The centre of the conic section $14x^2-4xy+11y^2-44x-58y+71=0$ is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $(2, 3)$

  2. $(2, -3)$

  3. $(-2, 3)$

  4. $(-2, -3)$

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Correct Option: A
Explanation:

Let $S\equiv 14x^2-4xy+11y^2-44x-58y+71$
To know centre of this hyperbola, $\cfrac{\partial S}{\partial x}=0$ and $\cfrac{\partial S}{\partial y}=0 $
$\Rightarrow \cfrac{\partial S}{\partial x}=28x-4y-44=0\Rightarrow 7x-y-11=0  ..(1)$
and $\cfrac{\partial S}{\partial y}=-4x+22y-58=0\Rightarrow 2x-11y+29=0  ..(2)$
Solving $(1)$ and $(2)$ we get required centre $(2,3)$
Hence, option 'A' is correct.

The vertices and the foci of a hyperbola are the points $\displaystyle \left ( \pm 5, 0 \right )$ and $\displaystyle \left ( \pm 7, 0 \right )$.Which of the following holds true?

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\displaystyle a^{2}\neq b^{2}$

  2. $a^2=b^2$

  3. $\dfrac{a^2}{b^2}=2$

  4. None of these

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Correct Option: A
Explanation:

$\displaystyle a= 5, ae = 7\Rightarrow a^{2}e^{2}= 49$
or $\displaystyle a^{2}\left ( 1+\frac{b^{2}}{a^{2}} \right )= 49$ 

Or 
$\displaystyle a^{2}+b^{2}=49$ 
$\displaystyle b^{2}= 24$ 
Since, $\displaystyle a^{2}\neq b^{2},$ hence hyperbola is not rectangular.

An ellipse intersects the hyperbola $2x^{2}-2y^{2}=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. equation of ellipse is $x^{2}+2y^{2}=2$

  2. the foci of ellipse are $\left ( \pm 1, 0 \right )$

  3. equation of ellipse is $x^{2}+2y^{2}=4$

  4. the foci of ellipse are $\left ( \pm \sqrt{2}, 0 \right )$

Show answer
Correct Option: A,B
Explanation:

Eccentricity of the hyperbola is $\sqrt{2}$ as it is a rectangular hyperbola. 

So eccentricity $e$ of the ellipse is $\dfrac1{\sqrt{2}}$
Let the equation of the ellipse be $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ where $b^{2}=a^{2}\left ( 1-e^{2} \right )=\dfrac{a^{2}}2\Rightarrow a^{2}=2b^{2}$
So equation of the ellipse is $x^{2}+2y^{2}=a^{2}$
Let $\left ( x _{1}, y _{1} \right )$ be a point of intersection of the ellipse and the hyperbola
Then $2x _{1}^{2}-2y _{1}^{2}=1$ and $x _{1}^{2}+2y _{1}^{2}=a^{2}$          (1)
Equations of the tangents at $\left ( x _{1}, y _{1} \right )$ to the two conics are
   $2xx _{1}-2yy _{1}=1$ and $xx _{1}+2yy _{1}=a^{2}$
Since the two conics intersect orthogonally
$\displaystyle \left ( \frac{x _{1}}{y _{1}} \right )\left ( -\frac{x _{1}}{2y _{1}} \right )=-1\Rightarrow x _{1}^{2}=2y _{1}^{2}$
And from (1) we get $x _{1}^{2}=1$, $a^{2}=2$.
Hence the equation of the ellipse is $x^{2}+2y^{2}=2$ and its focus is
   $\displaystyle \left ( \pm ae, 0 \right )=\left ( \pm \sqrt{2}\times \frac{1}{\sqrt{2}}, 0 \right )=\left ( \pm 1, 0 \right )$