Tag: hyperbola

Questions Related to hyperbola

Find the range of $p$ such that no perpendicular tangents can be drawn to the hyperbola $\dfrac{x^2}{(-p^2 + 6p + 5)} - \dfrac{y^2}{(-p - 3)} = 1$, i.e. the director circle of the given hyperbola is imaginary.

  1. $R - [-1 , 8]$

  2. $(5 , 6)$

  3. $(3 , 4)$

  4. $(-7 , 4)$


Correct Option: A
Explanation:

The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola.

Equation of the director circle of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2  b^2$ i.e. a circle whose center is origin and radius is $(a^2  b^2)$.

Hence, for the director circle to be imaginary, $a^2 < b^2$ .

$-p^2 + 6p + 5 < -p - 3$ ---> $p^2 -7p -8 > 0$ - $(p-  8)(p + 1) > 0$ $\Rightarrow$ $p$ lies in $R - [-1 , 8]$. Hence, option (A).

For the hyperbola $\dfrac{x^2}{49}-\dfrac{y^2}{25}=1$, the equation of auxillary circle is

  1. $x^2+y^2=49$

  2. $x^2+y^2=25$

  3. $x^2+y^2=10$

  4. $x^2+y^2=10074$


Correct Option: A
Explanation:

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

Here the given hyperbola is $\dfrac{x^2}{49}-\dfrac{y^2}{25}=1$, which is similar to standard form of the hyperbola.

Here $a = 7$ and $b = 5$

The equation of Auxiliary circle for the given hyperbola will also be $x^2 + y^2 = (7)^2$

$\Rightarrow x^2 + y^2 = 49$

So the correct option is $A$

The radius of the director circle of the ellipse $9{x^2} + 25{y^2} - 18x - 100y - 116 = 0$ is 

  1. $\sqrt {34} \,$

  2. $\sqrt {29} \,\,\,$

  3. 5

  4. 8


Correct Option: A

The  equation of  auxillary circle of  $\dfrac{x^2}{64}-\dfrac{y^2}{36}=1$ is

  1. $x^2+y^2=100$

  2. $x^2+y^2=50$

  3. $x^2+y^2=64$

  4. $x^2+y^2=36$


Correct Option: C
Explanation:

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

Here the given hyperbola is $\dfrac{x^2}{64}-\dfrac{y^2}{36}=1$, which is similar to standard form of the hyperbola.

Here $a = 8$ and $b = 6$

The equation of Auxiliary circle for the given hyperbola will also be $x^2 + y^2 = (8)^2$

$\Rightarrow x^2 + y^2 = 64$

So the correct option is $C$

For the hyperbola $\dfrac{x^2}{15}-\dfrac{y^2}{10}=1$, the equation of auxillary circle is

  1. $x^2+y^2=15$

  2. $x^2+y^2=10$

  3. $x^2+y^2=35$

  4. $x^2+y^2=5$


Correct Option: A
Explanation:

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

Here the given hyperbola is $\dfrac{x^2}{15}-\dfrac{y^2}{10}=1$, which is similar to standard form of the hyperbola.

Here $a = \sqrt{15}$ and $b = \sqrt{10}$

The equation of Auxiliary circle for the given hyperbola will also be $x^2 + y^2 = (\sqrt{15})^2$

$\Rightarrow x^2 + y^2 = 15$

So the correct option is $A$

If ${e _1}$and ${e _2}$ are the eccentricities of the hyperbolas $xy = 9$ and ${x^2} - {y^2} = 25$ ,then( ${e _1}$,${e _2}$) lie on a circle ${C _1}$with centre origin then the ${(radius)^2}$ of the director circle of ${C _1}$is

  1. $2$

  2. $4$

  3. $8$

  4. none


Correct Option: A

 The equation of auxillary circle is $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$

  1. $x^2+y^2=16$

  2. $x^2+y^2=32$

  3. $x^2+y^2=25$

  4. $x^2+y^2=41$


Correct Option: C
Explanation:

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,


The circle drawn taken major axis as a diameter also called the Auxiliary circle of the Hyperbola, will have a diameter of $2a$, equal to the length of major axis and center same as center of Hyperbola.

Hence equation of Auxiliary circle of any standard Hyperbola will be $x^2+y^2=a^2$

Here the given hyperbola is $\dfrac{x^2}{25}-\dfrac{y^2}{16}=1$, which is similar to standard form of the hyperbola.

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

The equation of Auxiliary circle for the given hyperbola will also be $x^2 + y^2 = (5)^2$

$\Rightarrow x^2 + y^2 = 25$

So the correct option is $C$

If the chords of contact of tangents drawn from $P$ to the hyperbola $x^2 - y^2 = a^2$ and its auxiliary circle are at right angle, then $P$ lies on :

  1. $x^2 - y^2 = 3a^2$

  2. $x^2 - y^2 = 2a^2$

  3. $x^2 - y^2 = 0$

  4. $x^2 - y^2 = 1$


Correct Option: C
Explanation:

Let $P$ be $(h,k)$
Now Chord of contact of tangent from $P$ to the hyperbola $x^2-y^2=a^2$ is,
$T =0\Rightarrow hx -ky = a^2$ (i)
And director circle of given hyperbola is, $x^2+y^2=a^2$
Thus equation of chord of contact to this circle from P is, $hx+ky = a^2$ (ii)
Now given line (i) and (ii) are perpendicular,
$\Rightarrow \cfrac{h}{k}\times \cfrac{-h}{k}=-1\Rightarrow h^2=k^2$
Hence locus of $P$ is given by, $x^2-y^2=0$ 

If the circle $x^2\, +\, y^2\, =\, a^2$ intersects the hyperbola $xy\, =\, c^2$ in four points $P\, (x _1,\, y _1),\, Q(x _2,\, y _2),\, R(x _3,\, y _3),\, S(x _4,\, y _4)$, then -

  1. $X _1\, +\, X _2\, +\, X _3\, +\, X _4\, =\, 0$

  2. $Y _1\, +\, Y _2\, +\, Y _3\, +\, Y _4\, =\,0$

  3. $X _1\, X _2\, X _3\, X _4\, =\, c^4$

  4. $Y _1\,Y _2\,Y _3\,Y _4\, =\, c^4$


Correct Option: A,B,C,D
Explanation:

Since, the circle $x^2\, +\, y^2\, =\, a^2$ intersects the hyperbola $xy\, =\, c^2$
Therefore, $x^2+\dfrac{c^4}{x^2}=a^2$
$\Rightarrow x^4-a^2x^2+c^4=0$
now sum of the roots: $x _1+x _2+x _3+x _4=0$
and product of the roots $x _1x _2x _3x _4=c^4$
Repeat the same for $y$

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

  1. $a-b$

  2. $\sqrt{a-b}$

  3. $\sqrt{a^{2}-b^{2}}$

  4. $\sqrt{a^{2}+b^{2}}$


Correct Option: A