Tag: hyperbola

Questions Related to hyperbola

Equation of the transverse and conjugate axis of a hyperbola are respectively $x+2y-3=0$, $2x-y+4=0$ and their respectively lengths are $\sqrt {2}$ and $\cfrac { 2 }{ \sqrt { 3 }  } $ then answer the following 
Equation of one of the directrix is

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

  1. $2x-y+4+\sqrt { \cfrac { 3 }{ 2 } } =0\quad $

  2. $x+2y+4-\sqrt { \cfrac { 2 }{ 3 } } =0$

  3. $2x-y=\sqrt { \cfrac { 3 }{ 2 } } $

  4. $2x-y+4+\sqrt { \cfrac { 3 }{ 2 } } =\sqrt { 3 } $

Show answer
Correct Option: A

The equation of a hyperbola is given in its standard form as $16x^2-9y^2=144$.Coordinates of foci is

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

  1. $(0, \pm 1)$

  2. $(0, \pm 1, 0)$

  3. $(\pm 5, 0)$

  4. $(0, \pm 5)$

Show answer
Correct Option: D
Explanation:

$Given\quad :\quad 16{ x }^{ 2 }−9{ y }^{ 2 }=144\quad \quad \quad \ Or,\quad \frac { { x }^{ 2 } }{ 9 } -\frac { { y }^{ 2 } }{ 16 } =1\ We\quad know,\ be=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \quad \quad \quad \quad \quad \because (b>a)\ Or,\quad be=\sqrt { 9+16 } \ Or,\quad be=\pm 5\ \therefore \quad Focii\quad is\quad (0,\pm 5)\quad $


Option [D]

Hyperbola $\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{3}=1$ of eccentricity $e$ is confocal with the ellipse $\dfrac{{x}^{2}}{8}+\dfrac{{y}^{2}}{4}=1$. Let $A$, $B$, $C$ & $D$ are points of intersection of hyperbola & ellipse, then-

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

  1. $e=\dfrac{5}{2}$

  2. $e=2$

  3. $A$, $B$, $C$, $D$ are concyclic points

  4. Number of common tangents of hyperbola & ellipse is $2$

Show answer
Correct Option: A

The foci of hyperbola $9x^2-16y^2+18x+32y=151$ are 

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

  1. $(-4,1),(6,1)$

  2. $(-11,2),(-6,1)$

  3. $(4,1),(-6,1)$

  4. $(2,1),(1,-6)$

Show answer
Correct Option: C
Explanation:
$9{x}^{2}−16{y}^{2}+18x+32y=151$
$\left(9{x}^{2}+18x\right)-\left(16{y}^{2}-32y\right)=151$
$\Rightarrow \left({\left(3x\right)}^{2}+2\times 3x\times 3+{3}^{2}-{3}^{2}\right)-\left({\left(4y\right)}^{2}-2\times 4y\times 4+{4}^{2}-{4}^{2}\right)=151$ by completing the square method
$\Rightarrow {\left(3x+3\right)}^{2}-9-{\left(4y-4\right)}^{2}+16=151$
$\Rightarrow {\left(3x+3\right)}^{2}-{\left(4y-4\right)}^{2}=151-7$
$\Rightarrow {\left(3x+3\right)}^{2}-{\left(4y-4\right)}^{2}=144$
$\Rightarrow 9{\left(x+1\right)}^{2}-16{\left(y-1\right)}^{2}=144$
$\Rightarrow \dfrac{9{\left(x+1\right)}^{2}}{144}-\dfrac{16{\left(y-1\right)}^{2}}{144}=1$ by dividing both sides by $144$
$\Rightarrow \dfrac{{\left(x+1\right)}^{2}}{16}-\dfrac{{\left(y-1\right)}^{2}}{9}=1$ is the equation of the horizontal hyperbola.
center$=\left(-1,1\right)$
We have $a=4$ and $b=3$
${c}^{2}={a}^{2}+{b}^{2}={4}^{2}+{3}^{2}=16+9=25$
$\therefore c=\sqrt{25}=\pm 5$
Foci$=\left(-1\pm 5, 1\right)$
$\therefore$Foci$=\left(-1+5,1\right)$ and $\left(-1-5,1\right)$
Hence Foci$=\left(4,1\right)$ and $\left(-6,1\right)$

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 $=12$

Show answer
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. $(-2, -3)$

Show answer
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

Show answer
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
   $\displaystyle \left ( \pm ae, 0 \right )=\left ( \pm \sqrt{2}\times \frac{1}{\sqrt{2}}, 0 \right )=\left ( \pm 1, 0 \right )$