Tag: hyperbola

Questions Related to hyperbola

If a line intersect a hyperbola at $(-2,-6)$ and $(4,2)$ and one of the asymtote at $(1,-2)$, then the centre of the hyperbola is

  1. $(7,6)$

  2. $(1,-2)$

  3. $(10,10)$

  4. $(-5,-10)$


Correct Option: A

Let product of distances of any point hyperbola (x+y-1) (x-y+3)= 60 to its asymptotes is 'K' then K is divisible by

  1. 2

  2. 3

  3. 4

  4. 5


Correct Option: A

If the cordinate of any point p on the hyperbola $9{x^2} - 16{y^2} = 144$ is produced to cut the asymptotes in the points Q and R. Then the product PQ.PR equals to:

  1. $9$

  2. $\dfrac{12}{5} $

  3. $\dfrac{144}{25}$

  4. $7$


Correct Option: C
Explanation:
$\dfrac{x^{2}}{16}-\dfrac{y^{2}}{9}=1$

Asymplote is $y=\pm\dfrac{3}{4}x$

Let us take $4y=3x, 4y=-3x$

consider a parametric point $(4\sec\theta, 3\tan\theta)$ on the parabola

$Q$ is intersection with $4y=3x$

then $PQ=\left|\dfrac{12\sec\theta-12\tan\theta}{\sqrt{4^{2}+3^{2}}}\right|$

$PQ=\left|\dfrac{12}{5}(\sec \theta-\tan\theta)\right|$

$R$ is intersection with $4y=-3x$

then $PR=\left|\dfrac{12\sec\theta+12\tan\theta}{\sqrt{4^{2}+3^{2}}}\right|$

$PQ.PR=\dfrac{144}{25}(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)$

$=\dfrac{144}{25}(\sec^{2}\theta+\tan^{2}\theta)=\dfrac{144}{25}(1)$

`e`$=\dfrac{144}{25}$

The points of intersection of asymptotes with directrices lies on

  1. Auxillary circle

  2. Director circle

  3. Transverse axis

  4. Conjugate axis


Correct Option: A
Explanation:
  • Asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis.
  • A perpendicular drawn from the foci on either asymptote meet it in the same points as the corresponding directrix and the common points of intersection lie on the auxiliary circle.

The area of the triangle formed by the asymptotes and any tangent to the hyperbola ${x}^{2}-{y}^{2}={a}^{2}$ is 

  1. ${4a}^{2}$

  2. ${3a}^{2}$

  3. ${2a}^{2}$

  4. ${a}^{2}$


Correct Option: A

If foci of hyperbola lie on $y=x$ and one of the asymptote is $y=2x$, then equation of the hyperbola, given that is passes through $(3, 4)$ is :

  1. $x^2-y^2-\dfrac {5}{2}xy+5=0$

  2. $2x^2-2y^2+5xy+5=0$

  3. $2x^2+2y^2-5xy+10=0$

  4. None of these


Correct Option: C
Explanation:

Foci of hyperbola lie on $y=x$.
So, the equation of transverse axis is $y-x=0$.
Transverse axis of hyperbola bisects the asymptote
$\Rightarrow$ equation of other asymptote is $y=\dfrac{x}{2}$
or,$x=2y$
$\Rightarrow$ Equation of hyperbola is $(y-2x)(x-2y)+k=0$
Since, it passes through $(3, 4)$
$\Rightarrow k=-10$
Hence, required equation is
$2x^2+2y^2-5xy+10=0$

The combined equation of the asymptotes of the hyperbola $2{x}^{2}+5xy+2{y}^{2}+4x+5y=0$ is

  1. $2{x}^{2}+5xy+2{y}^{2}+4x+5y+2=0$

  2. $2{x}^{2}+5xy+2{y}^{2}+4x+5y-2=0$

  3. $2{x}^{2}+5xy+2{y}^{2}=0$

  4. None of these


Correct Option: A

The ordinate of any point P on the hyperbola, given by  $25x^2-16y^2=400$, is produced to cut its asymptotes in the points Q and R, then $QP.PR=5.$

  1. True

  2. False


Correct Option: B

If the x-y+4=0 and x+y+2=0 are asymptotes of a hyperbola , the its center is 

  1. (-3,1)

  2. (3,1)

  3. (-3,-1)

  4. (3,-1)


Correct Option: A

A chord $AB$ which bisected at $(1,1)$ is drawn to the hyperbola $7x^{2}+8xy-y^{2}-4=0$ with centre $C$. which intersects its asymptotes in $E$ and $F$. If equation of circumcricel of $\triangle CEF$ is $x^{2}+y^{2}-ax-by+c=0$, then value of $\dfrac{23(a-b+c)}{12}$ is equal to 

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A