Tag: director circle and auxiliary circle of a hyperbola

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Questions Related to director circle and auxiliary circle of a hyperbola

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The radius of  director circle of hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

  1. $a$
  2. $b$
  3. $\sqrt{a^2+b^2}$
  4. $\sqrt{a^2-b^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Hence the Director circle is a circle whose centre is same as centre of the hyperbola and the radius is $\sqrt{a^2 - b^2}$

So correct option is $D$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The equation of director circle of $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is:

  1. Imaginary if $a < b$
  2. Imaginary if $a>b$
  3. Point circle if $a=b$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
The director circle of an hyperbola circumscribes the minimum bounding box of the hyperbola.it has the same center as the hyperbola, with radius $\sqrt { { a }^{ 2 }-{ b }^{ 2 } } $ where $a$ and $b$ are the semi-major axis and semi-minor axis of the hyperbola.
equation of circle is 
$x^2+y^2=a^2-b^2$
$a<b$ then circle radius is negative and circle is imaginary 
$a=b$ then circle radius is zero and circle is point circle
Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The director circle intersects its hyperbola in _______ number of points.

  1. zero

  2. two

  3. three

  4. four
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the equation of director circle is given by:


$x^2 + y^2 = a^2-b^2$

So we can see the center of the director circle of the standard hyperbola is same as the hyperbola. the radius of director circle is $\sqrt{a^2-b^2}$

As $a^2>0$, $a^2 > a^2 -b^2$

or $a > \sqrt{a^2-b^2}$

So we can see the radius of director circle is less the vertex of the hyperbola. Hence with a center same as the hyperbola, the director circle doesn't cut the hyperbola at any point. or we can say it's just smaller than the Auxiliary circle of the hyperbola which touches the vertices of the hyperbola. 

Hence the correct option is $A$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The radius of the director circle of the hyperbola $\dfrac{x^2}{a(a+4b)}-\dfrac{y^2}{b(2a-b)}=1; 2a > b > 0$ is: 

  1. $a^2+b^2+4ab$
  2. $a+b$
  3. $a^2+b^2+2ab$
  4. $2(a+b)^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} equation\, \, of\, \, director\, \, circle\, \, { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+4ab-\left( { 2ab-{ b^{ 2 } } } \right)  \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } }+2ab \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }={ \left( { a+b } \right) ^{ 2 } } \ Then,\, \, Radians\, \, of\, \, circle\, \, is\left( { a+b } \right)  \end{array}$

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The diametre of director circle of hyperbola $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$ 

  1. $3$
  2. $9$
  3. $6$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Here the given hyperbola is $\dfrac{x^2}{25} -\dfrac {y^2}{16} = 1$,

Here $a =5$ and $b =4$

So equation of the director circle will be $x^2 + y^2 = (5)^2 - (4)^2$

$\rightarrow x^2 + y^2 = 9$

Hence the radius of the director circle is $\sqrt9 = 3$. So the diameter will be $6$.

So correct option is $C$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

 The  equation of director circle of hyperbola is $\dfrac{x^2}{36}-\dfrac{y^2}{25}=1$ is

  1. $x^2+y^2=4$
  2. $x^2+y^2=11$
  3. $x^2-y^2=4$
  4. $x^2+y^2=61$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Here the given hyperbola is $\dfrac{x^2}{36} -\dfrac {y^2}{25} = 1$,

Here $a =6$ and $b =5$

So equation of the director circle will be $x^2 + y^2 = (6)^2 - (5)^2$

$\Rightarrow x^2 + y^2 = 11$

Correct option is $B$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

Point P is on the orthogonal hyperbola $x^2 - y^2 = a^2$. Point P' is the perpendicular projection of P on the x-axis. Then, $|PP'|^2$ is equal to the power of point P' relative to which circle?

  1. $x^2 + y^2 = a^2$
  2. $x^2 + y^2 = a^2 + b^2$

  3. Director circle

  4. Auxiliary circle
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

$P(a sec(t) , a tan(t))$ and $P'(a sec(t) , 0)$
$|PP'| = a^2 tan^2(t)$
The power of point P' relative to a circle $x^2 + y^2 = a^2$ is :

$(asec(t) - 0)^2 + (0-0)^2 - a^2$   (power of the point w.r.t. circle)

$= a^2sec^2(t) - a^2 = a^2(sec^2(t)-1) = a^2tan^2(t)$

Hence, the correct options are A and D.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The pole of the line $lx + my + n = 0$ with respect to the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, is

  1. $\displaystyle \left ( \frac{a^2 l}{n} , \frac{b^2 m}{n} \right )$
  2. $\displaystyle \left ( - \frac{a^2 l}{n} , \frac{b^2 m}{n} \right )$
  3. $\displaystyle \left ( \frac{a^2 l}{n} , -\frac{b^2 m}{n} \right )$
  4. $\displaystyle \left ( -\frac{a^2 l}{n} , -\frac{b^2 m}{n} \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the pole of the line

$lx+my+n=0$ with respect ot the hyperbola $\cfrac { { x }^{ 2 } }{ {

a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of the polar is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the pole of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { -y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { { b }^{ 2 }m }{ n } $
Hence the pole of the given line with respect of the given hyperbola is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The number of points from where a pair of perpendicular tangents can be drawn to the hyperbola, $ x^2 \sec^2\alpha-y^2 \cos ec^2\alpha=1, \alpha\in(0,\dfrac{\pi}4) $ are

  1. $0$
  2. $1$
  3. $2$

  4. infinite
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tangent equation to the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is 
$y=mx \pm \sqrt{a^2m^2-b^2}$
$\Rightarrow (y-mx)=\pm \sqrt{a^2m^2-b^2}$
On squaring both sides, we get
$\Rightarrow (y-mx)^2=(a^2m^2-b^2)$
$\Rightarrow (x^2-a^2)m^2-2xym+(y^2+b^2)=0$
Product of the slopes,$m _1m _2=\dfrac{(y^2+b^2)}{x^2-a^2}$
But given tangents are perpendicular to each other $\Rightarrow$ Their product of slopes equal to $-1.$
$\Rightarrow m _1m _2=-1$
$\Rightarrow \dfrac{(y^2+b^2)}{x^2-a^2}=-1$
$\Rightarrow x^2+y^2=(a^2-b^2)$
But given hyperbola equation as $\dfrac{x^2}{\cos\alpha^2}-\dfrac{y^2}{\sin\alpha^2}=1$
The required tangent equation is $x^2+y^2=\cos^2\alpha-\sin^2\alpha=\cos 2\alpha$
Since radius of circle is always greater than equal to zero.
$\Rightarrow \cos 2\alpha \geq 0$
But maximum value of $\cos$ is $1$.
$\Rightarrow 0 \leq \cos2\alpha \leq 1$
$\Rightarrow  \dfrac{\pi}{2} \leq 2\alpha \leq 0$
$\Rightarrow  \dfrac{\pi}{4} \leq \alpha \leq 0$
$\Rightarrow \alpha$has inifinite number of solutions.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The locus of the point of intersection of two perpendicular tangents to the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is

  1. Director circle
  2. $x^2 + y^2 = a^2$
  3. $x^2 + y^2 = a^2 - b^2$
  4. $x^2 + y^2 = a^2 + b^2$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Equation of any tangent in terms of slope $m$ is

$y = mx + (a^2m^2  b^2)$

It passes through $(h, k)$, so we have

$(k - mh)^2 = a^2m^2 - b^2$

So, $m^2(h^2 - a^2) - 2mhk + k^2 + b^2 = 0$

This is a quadratic in $m$

Let the slopes of tangents be $m _1$ and $m _2$.

then $m _1.m _2 = -1$.

So, $\dfrac{(k^2 + b^2)}{(h^2 - a^2)} = -1$

$(h^2 + k^2) = (a^2  b^2)$

Hence, the locus is $(x^2 + y^2) = (a^2  b^2)$ which is the director circle of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1.$