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 eccentricity of the hyperbola $16x^2-9y^2=1$ is

  1. $\dfrac{3}{5}$
  2. $\dfrac{5}{3}$
  3. $\dfrac{4}{5}$
  4. $\dfrac{5}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of hyperbola: $16x^2-9y^2=1$
can be written as $\cfrac{x^2}{\frac{1}{16}}+\cfrac{y^2}{\frac{1}{9}}=1$
$e^2=1+\cfrac{b^2}{a^2}=1+\cfrac{16}{9}=\cfrac{25}9$
or, $e=\cfrac53$
Hence, B is the correct option.
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

For hyperbola  $-\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$ distance between directrices is 

  1. $\dfrac{50}{\sqrt{41}}$
  2. $\dfrac{16}{\sqrt{41}}$
  3. $\dfrac{25}{\sqrt{41}}$
  4. $\dfrac{32}{\sqrt{41}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given equation is $-\dfrac {x^2}{16}+\dfrac {y^2}{25}=1$
$\Rightarrow \dfrac {y^2}{25}-\dfrac {x^2}{16}=1$
Distance between directrix is   $\dfrac{2a}{\sqrt{a^2+b^2}}$
Here $a=25, b=16$
So, answer is $= \dfrac{50}{\sqrt{41}}$.

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

For hyperbola  $-\dfrac{(x-1)^2}{3}+\dfrac{(y+2)^2}{16}=1$ centre is 

  1. $(-1,-2)$
  2. $(1,-1)$
  3. $(1,-2)$
  4. $(0,0)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given hyperbola is $-\dfrac{(x-1)^2}{3} + \dfrac{(y+2)^2}{16} = 1$ is a conjugate hyperbola.


It can be written as $\dfrac{(x-1)^2}{3} - \dfrac{(y+2)^2}{16} = -1$  ......$(1)$

Now let $x-1 = X$ and $y+2 = Y$

Putting values of $x-1$ and $y+2$ in eq. $(2)$ we get,

$\rightarrow  \ \dfrac{(X)^2}{3} - \dfrac{(Y)^2}{16} = -1$  ....$(2)$

We can see the eq$(2)$ is a standard form of conjugate hyperbola and it's center lies at origin $(0,0)$

So $X =0 , Y = 0$ is the center of the hyperbola given in eq.$(2)$

$\rightarrow$ $ X = x-1 = 0$ or $x = 1$

$\rightarrow$ $ Y = y+2 = 0$ or $y = -2$

Hence center of the given hyperbola in eq. $(1)$ is $(1,-2)$. So correct option is $C$.

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

The equation of the conjugate axis of the hyperbola $\dfrac {(y - 2)^{2}}{9} - \dfrac {(x + 3)^{2}}{16} = 1$ is

  1. $y = 2$
  2. $y = 6$
  3. $y = 8$
  4. $y = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation at the conjugate axis of the hyperbola

$\frac{{{{\left( {y - 2} \right)}^2}}}{9} - \frac{{{{\left( {x + 3} \right)}^2}}}{9} = 1$ 
There for,
      $y-2=0$
      $y=2$
 option (A) is correct answer

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

An ellipse and a hyperbola have the same principle axes. From a point on the ellipse, tangents are drawn to the hyperbola . then  the chord contact of these tangents touches the ellipse.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard property of conics where the chord of contact of tangents from a point on one conic to another conic touches the original conic under specific conditions.

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

The eccentricity of the conic represented by$2{x}^{2}+5xy+2{y}^{2}+11x-7y-4=0$ is

  1. $\dfrac {\sqrt {10}}{3}$
  2. $\dfrac {\sqrt {10}}{4}$
  3. $\dfrac {5}{4}$
  4. $\dfrac {3}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation represents a hyperbola. Calculating the eccentricity involves finding the angle between the asymptotes or using the general conic eccentricity formula. The result sqrt(10)/3 is the standard derivation for this specific equation.

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

The equation $\dfrac{x^{2}}{29 -p} + \dfrac{y^{2}}{4 -p} =1(p\neq4, 29)$ represents - 

  1. an ellipse if $p$ is any constant greater than $4$
  2. hyperbola if $p$ is any constant between $4$ and $29$.
  3. a rectanglar hyperbola is $p$ is any constant greater than $29$.
  4. no real curve is $p$ is less than $29$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of Hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

if p lies between $4$ and $29$ then coefficient of $y^2$  is negative and coefficient of $x^2$ is positive

Hence, it satisfies the equation of Hyperbola between $4$ and $29$
Multiple choice mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

Which of the following equations in parametric form can represent a hyperbolic profile, where $t$ is a parameter.

  1. $x=\dfrac { a }{ 2 } \left( t+\dfrac { 1 }{ t } \right)$ & $y=\dfrac{b}{2}\left( t-\dfrac { 1 }{ t } \right)$
  2. $\dfrac{tx}{a}-\dfrac{y}{b}+t=0$ & $\dfrac{x}{a}-\dfrac{ty}{b}-1=0$
  3. $x={e}^{t}+{e}^{-t}$ & $x={e}^{t}-{e}^{-t}$
  4. ${x}^{2}-6=2\cos{t}$ & ${y}^{2}+2=4{\cos}^{2}\dfrac{t}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(a)$ We have $x=\dfrac{a}{2}\left(t+\dfrac{1}{t}\right)$ and $y=\dfrac{b}{2}\left(t-\dfrac{1}{t}\right)$

$\Rightarrow\,\dfrac{2x}{a}=t+\dfrac{1}{t}$ and $\dfrac{2y}{b}=t-\dfrac{1}{t}$

$\Rightarrow\,{\left(\dfrac{2x}{a}\right)}^{2}={\left(t+\dfrac{1}{t}\right)}^{2}$ and ${\left(\dfrac{2y}{b}\right)}^{2}={\left(t-\dfrac{1}{t}\right)}^{2}$

$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}+2$ and $\dfrac{4{y}^{2}}{{b}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}-2$

$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}-\dfrac{4{y}^{2}}{{b}^{2}}={t}^{2}+\dfrac{1}{{t}^{2}}+2-{t}^{2}-\dfrac{1}{{t}^{2}}+2$

$\Rightarrow\,\dfrac{4{x}^{2}}{{a}^{2}}-\dfrac{4{y}^{2}}{{b}^{2}}=4$

$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ is in parametric form can represents a hyperbolic profile.

$(b)\dfrac{tx}{a}-\dfrac{y}{b}+t=0$ and $\dfrac{x}{a}-\dfrac{ty}{b}-1=0$

$\Rightarrow\,\dfrac{tx}{a}+t=\dfrac{y}{b}$ and $\dfrac{x}{a}-1=\dfrac{ty}{b}$

$\Rightarrow\,t\left(\dfrac{x}{a}+1\right)=\dfrac{y}{b}$ and $\dfrac{b}{y}\left(\dfrac{x}{a}-1\right)=t$

$\Rightarrow\,t\left(\dfrac{x+a}{a}\right)=\dfrac{y}{b}$ and $t=\dfrac{b}{y}\left(\dfrac{x-a}{a}\right)$

$\Rightarrow\,\dfrac{b}{y}\left(\dfrac{x-a}{a}\right)\left(\dfrac{x+a}{a}\right)=\dfrac{y}{b}$ 

$\Rightarrow\,\dfrac{b}{y}\left(\dfrac{{x}^{2}-{a}^{2}}{{a}^{2}}\right)=\dfrac{y}{b}$ 

$\Rightarrow\,\dfrac{{x}^{2}-{a}^{2}}{{a}^{2}}=\dfrac{{y}^{2}}{{b}^{2}}$ 

$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-1=\dfrac{{y}^{2}}{{b}^{2}}$ 

$\Rightarrow\,\dfrac{{x}^{2}}{{a}^{2}}-\dfrac{{y}^{2}}{{b}^{2}}=1$ is in parametric form can represents a hyperbolic profile.

$(c)x={e}^{t}+{e}^{-t}$ and $x={e}^{t}-{e}^{-t}$

$\Rightarrow\,{x}^{2}={e}^{2t}+{e}^{-2t}+2{e}^{t}{e}^{-t}$ and $ {x}^{2}={e}^{2t}+{e}^{-2t}-2{e}^{t}{e}^{-t}$

$\Rightarrow\,{x}^{2}={e}^{2t}+{e}^{-2t}+2$ and ${x}^{2}={e}^{2t}+{e}^{-2t}-2$

$\Rightarrow\,{x}^{2}-{x}^{2}={e}^{2t}+{e}^{-2t}+2-{e}^{2t}-{e}^{-2t}+2=4$ is not in parametric form can represents a hyperbolic profile.

$(d){x}^{2}-6=2\cos{t}$ and ${y}^{2}+2=4{\cos}^{2}{\dfrac{t}{2}}$

$\Rightarrow\,{x}^{2}=2\cos{t}+6$ and ${y}^{2}=4{\cos}^{2}{\dfrac{t}{2}}-2$

$\Rightarrow\,{x}^{2}-{y}^{2}=2\cos{t}+6-4{\cos}^{2}{\dfrac{t}{2}}+2$ 

$\Rightarrow\,{x}^{2}-{y}^{2}=2\left(2{\cos}^{2}{\dfrac{t}{2}}-1\right)+8-4{\cos}^{2}{\dfrac{t}{2}}$ 

$\Rightarrow\,{x}^{2}-{y}^{2}=4{\cos}^{2}{\dfrac{t}{2}}-2+8-4{\cos}^{2}{\dfrac{t}{2}}$ 

$\Rightarrow\,{x}^{2}-{y}^{2}=6$ is in parametric form can represents a hyperbolic profile.


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

The transverse axis of a hyperbola is of length $2a$ and a vertex divides the segment of the axis between the centre and the corresponding focus in the ratio $2:1$. The equation of the hyperbola is

  1. $4x^2-5y^2=4a^2$
  2. $4x^2-5y^2=5a^2$
  3. $5x^2-4y^2=4a^2$
  4. $5x^2-4y^2=5a^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have given  $\cfrac{a}{ae-a}=2\Rightarrow e=\cfrac{3}{2}$
Using $e^2=1+\cfrac{b^2}{a^2}$
$\cfrac{9}{4}=1+\cfrac{b^2}{a^2}\Rightarrow b^2=\cfrac{5}{4}a^2$
Hence required hyperbola is $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$
$\cfrac{x^2}{a^2}-\cfrac{y^2}{\frac{5}{4}a^2}=1$
$\Rightarrow 5x^2-4y^2=5a^2$
Hence option 'D' 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 $2x + y = 1$,corresponding focus is $(1, 1)$ and eccentricity $\sqrt { 3 }$, is given by 

  1. $7 x ^ { 2 } + 12 x y - 2 y ^ { 2 } - 2 x + 4 y - 7 = 0$
  2. $2 x ^ { 2 } + 12 x y - 7 y ^ { 2 } - 2 x + 14 y - 7 = 0$
  3. $7 x ^ { 2 } - 12 x y + 2 y ^ { 2 } - 2 x + 14 y - 22 = 0$
  4. $7 x ^ { 2 } + 12 x y - 2 y ^ { 2 } - 2 x - 14 y - 22 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let$ P(x, y)$ is any point on the hyperbola.
given, focus of parabola is $S(1,1)$.
equation of directrix is $2x + y = 1$
From P draw PM perpendicular to the directrix then $PM = (2x + y – 1)/√(2² + 1²) = (2x + y – 1)/√5$
Also from the definition of the hyperbola, we have
$SP/PM = e ⇒ SP = ePM$
$⇒ √{(x–1)² + (y–1)²} = √3{(2x + y – 1)/√5}$
$⇒ (x – 1)² + (y – 1)² = 3 (2x + y – 1)²/5$
$⇒ 5[(x² – 2x + 1) + (y² –2y + 1)] = 3(4x² + y² + 1 + 4xy – 4x – 2y)$
$⇒5x² - 10x + 5 + 5y² - 10y + 5 = 12x² + 3y² + 3 + 12xy - 12x - 6y $
$⇒7x² + 2y² + 12xy - 2x + 4y - 7 = 0$
hence, equation of hyperbola is $7x² - 2y² + 12xy - 2x + 4y - 7 = 0$