Tag: hyperbola

Quizzes (10) Questions (269)

Questions Related to hyperbola

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

  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 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The directrix of a hyperbola is given by the line parallel to the conjugate axis at a distance a/e from the center. Given the axes and lengths, one can determine the specific equation for the directrix.

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

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

  1. $(0, \pm 1)$
  2. $(0, \pm 1, 0)$
  3. $(\pm 5, 0)$
  4. $(0, \pm 5)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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]

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

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-

  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 $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

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

  1. $(-4,1),(6,1)$
  2. $(-11,2),(-6,1)$
  3. $(4,1),(-6,1)$
  4. $(2,1),(1,-6)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\left(-6,1\right)$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of 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 :

  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 $=12$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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 $\displaystyle=\frac{2b^2}{a}=12$

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

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

  1. $(2, 3)$
  2. $(2, -3)$
  3. $(-2, 3)$
  4. $(-2, -3)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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?

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

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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 $\left ( \pm \sqrt{2}, 0 \right )$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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
   $\displaystyle \left ( \pm ae, 0 \right )=\left ( \pm \sqrt{2}\times \frac{1}{\sqrt{2}}, 0 \right )=\left ( \pm 1, 0 \right )$