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

The equation of the hyperbola whose foci are $(8,3)$ and $(0,3)$ and eccentricity$=\cfrac { 4 }{ 3 } $ is

  1. $ 7{\left(x-4 \right ) }^{2} -9{\left(y-3 \right) }^{2}=63$
  2. ${ 7x }^{ 2 }-{ 9y }^{ 2 }=63$
  3. $ 9{ \left( x-4 \right) }^{ 2 }-9{ \left( y-3 \right) }^{ 2 }=63$
  4. $7{ \left( x+4 \right) }^{ 2 }-9{ \left( y+3 \right) }^{ 2 }=63$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The centre of the hyperbola is the mid-point of the line joining the two
foci. So, the coordinates of the centre are $\left( \cfrac { 8+0 }{ 2 },\cfrac { 3+3 }{ 2 }  \right) \quad $ i.e $(4,3)$
Let $2a,2b$ be the length of the transverse and conjugate axes and let $e$ be the
eccentricity. Then, the equation of the hyperbola is
$\cfrac { {\left( x-4 \right)  }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { \left( y-3 \right)  }^{ 2 } }{ { b }^{ 2 } } =1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Distance between the two foci$ = 2ae$
$\Rightarrow \sqrt { { \left( 8-0 \right)  }^{ 2 }+{ \left( 3-3 \right)  }^{ 2 } } =2ae$
$\Rightarrow ae=4\Rightarrow a=3$
$\therefore\quad { b }^{ 2 }={ a }^{ 2 }\left( { e }^{ 2 }-1 \right) \Rightarrow {b }^{ 2 }=9\left( \cfrac { 16 }{ 9 } -1 \right) =7$
Substituting the value of $a$ and $b$ in $(i)$, we find that the equation of the hyperbola is
$\cfrac{ { \left( x-4 \right)  }^{ 2 } }{ 9 } +\cfrac { { \left( y-3 \right)  }^{ 2 } }{ 7 } =1\quad or\quad 7{ \left( x-4 \right)  }^{ 2 }-9{ \left( y-3 \right)  }^{ 2 }=63$
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 equation of the hyperbola whose directrix is $x + 2y = 1$, focus is $(2, 1)$ and eccentricity $2$ is

  1. $x^2 + 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
  2. $x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
  3. $x^2 - 4 xy - y^2 - 12 x + 6y + 21 = 0$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-2)^2+(y-1)^2=2^2\left(\cfrac{x+2y-1}{\sqrt{5}}\right)^2$
$5(x^2+y^2-4x-2y+5)=4(x^2+4y^2+1+4xy-2x-4y)$
$\Rightarrow x^2 - 16 xy - 11y^2 - 12 x + 6y + 21 = 0$
Hence, option 'B' is correct.

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

If ${ e } _{ 1 }$ is the eccentricity of the ellipse $\cfrac { { x }^{ 2 } }{ 16 } +\cfrac { { y }^{ 2 } }{ 25 } =1$ and ${ e } _{ 2 }$ is the eccentricity of the hyperbola passing through the foci of the ellipse and ${ e } _{ 1 }.{ e } _{ 2 }=1$, then the equation of the hyperbola, is :

  1. $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 16 } =1$
  2. $\cfrac { { x }^{ 2 } }{ 16 } -\cfrac { { y }^{ 2 } }{ 9 } =-1$
  3. $\cfrac { { x }^{ 2 } }{ 9 } -\cfrac { { y }^{ 2 } }{ 25 } =1$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have ${ e } _{ 1 }=\sqrt { 1-\cfrac { 16 }{ 25 }  } =\cfrac { 3 }{ 5 } $
$\because \quad { e } _{ 1 }{ e } _{ 2 }=1\Rightarrow { e } _{ 2 }=\cfrac { 5 }{ 3 } $
Clearly y-axis is transverse axis of the ellipse.
Thus, coordinates of foci of the ellipse are $(0,\pm b e _1)$ or $\left( 0,\pm 3 \right) $.
Let the equation of hyperbola is, $\dfrac{y^2}{b^2}-\cfrac{x^2}{a^2}=1$ ..... $(1)$
Since, hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also $a^2=b^2(e^2-1)=9(25/9-1)=16$
Therefore, the required hyperbola is, $\cfrac{x^2}{16}-\cfrac{y^2}{9}=-1$
Hence, option 'B' is correct.

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

Equation of the hyperbola whose vertices are at ($\pm3, 0$) and focii at ($\pm5, 0$) is

  1. $16x^2 - 9y^2 = 144$
  2. $9x^2 - 16y^2 = 144$
  3. $25x^2 - 9y^2 = 255$
  4. $9x^2 - 25y^2 = 81$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $a=3$ and $ae=5\Rightarrow e=\cfrac{5}{3}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{25}{9}=1+\cfrac{b^2}{9}\Rightarrow b^2=16$
Therefore required hyperbola is, $\cfrac{x^2}{9}-\cfrac{y^2}{16}=1$
$\Rightarrow 16x^2-9y^2=144$
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 equation of the conic with focus at $(1, -1)$, directrix along $x - y + 1= 0$ and with eccentricity $\sqrt{2}$ is

  1. $x^2 - y^2 = 1$
  2. $xy = 1$
  3. $2xy - 4x + 4y + 1 = 0$
  4. $2xy + 4x - 4y - 1 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Definition of hyperbola
$PS^2=e^2\cdot PM^2$
$(x-1)^2+(y+1)^2=2\left(\cfrac{x-y+1}{\sqrt{2}}\right)^2$
$(x^2+y^2-2x+2y+2)=(x^2+y^2+1-2xy+2x-2y)$
$\Rightarrow 2 xy - 4 x + 4y + 1 = 0$
Hence, option 'C' is correct.

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

The tangent of a point $P$ on the hyperbola $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$ passes through the point $(0,\ -b)$ and the normal at $P$ pases through the point $(2a\sqrt {2},\ 0)$. Then the eccentricity of the hyperbola is   

  1. $2$
  2. $\sqrt {2}$
  3. $3$
  4. $\sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the condition that the tangent passes through (0, -b) and the normal through (2a*sqrt(2), 0) for a point P on the hyperbola, one can derive the eccentricity e = 2.

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

Find the equation of the hyperbola whose directrix is $2x+y=1$, focus $(1,2)$ and eccentricity $\sqrt{3}$

  1. $7x^2-2y^2 +12xy-2x+14y-22=0$
  2. $7x^2-2y^2 +2xy-2x+14y-22=0$
  3. $7x^2-2y^2 +xy-14x+2y-22=0$

  4. none of above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the definition of a hyperbola (distance from focus = e * distance from directrix), the equation is (x-1)^2 + (y-2)^2 = 3 * (2x+y-1)^2 / (2^2 + 1^2). Expanding this yields the provided equation.

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

Eccentricity of the hyperbola satisfying the differential equation $2xy\dfrac{dy}{dx}=x^2+y^2$ and passing through $(2,1)$ is

  1. $\sqrt2$
  2. $2\sqrt2$
  3. $3\sqrt2$
  4. $5\sqrt2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the differential equation 2xy dy/dx = x^2 + y^2 leads to the hyperbola x^2 - y^2 = c. Passing through (2, 1) gives 4 - 1 = c, so c = 3. The hyperbola is x^2 - y^2 = 3. For a rectangular hyperbola, e = sqrt(2).

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

The ecentricity of the hyperbola passing through the origin and whose asymptotes are given by straight lines $y=3x-1$ and $x+3y=3$, is

  1. $\sqrt{2}$
  2. $3$
  3. $2\sqrt{2}$
  4. $\dfrac{3}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The asymptotes are 3x - y - 1 = 0 and x + 3y - 3 = 0. The angle between them is 90 degrees because the product of their slopes is -1 (3 * -1/3 = -1). A hyperbola with perpendicular asymptotes is a rectangular hyperbola, which has an eccentricity of sqrt(2).