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

Find the equation to the hyperbola of given transverse xis (2a) whose vertex bisects the distance between the centre and the focus

  1. $3x^2-2y^2=12a^2$
  2. $3x^2-y^2=a^2$
  3. $3x^2-y^2=3a^2$
  4. $3x^2-y^2=2a^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given transverse axis is 2a, so the center is at the origin and vertex is at (a, 0). The focus is at (ae, 0). Since the vertex bisects the distance between the center and the focus, a = ae / 2, which means e = 2. Using b^2 = a^2(e^2 - 1) gives b^2 = 3a^2, yielding the hyperbola equation 3x^2 - y^2 = 3a^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).