Tag: hyperbola

Questions Related to hyperbola

If the tangent at the point $(h, k)$ to the hyperbola $\dfrac{x^2}{a^2}\, -\, \dfrac{y^2}{b^2}\, =\, 1$ cuts the auxiliary circle in points whose ordinates are $y _1$ and $ y _2$, then  $\dfrac{1}{y _1} + \dfrac{1}{y _2} =$.

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

  1. $\dfrac{4}{k}$

  2. $\dfrac{3}{k}$

  3. $\dfrac{2}{k}$

  4. None of these

Show answer
Correct Option: C
Explanation:

Equation of tangent of given hyperbola at point
$\displaystyle (h,\, k)\,$ is $\dfrac{hx}{a^2}\, -\, \dfrac{ky}{b^2}\, =\, 1$ ...(i)
Equation of auxillary circle is $x^2\, +\, y^2\, =\, a^2$ .....(ii)
From (i) and (ii)
$\displaystyle \left [ \left ( 1\, +\, \frac{ky}{b^2}\right ) \frac{a^2}{h}\right ]^2\, +\, y^2\, -\, a^2\, =\, 0$
$\Rightarrow\, y^2\, (k^2a^4\, +\, b^4h^2)\, +\, 2kb^2a^4y\, +\, b^4a^2\, (a^2\, -\, h^2)\, =\, 0$
Now  $\displaystyle \, \frac{y _1\, +\, y _2}{y _1y _2}\, =\, -\,

\frac{2kb^2a^4}{b^4a^2(a^2\, -\, h^2)}\, =\, \frac{-2ka^2}{b^2a^2 \left (

1\, -\, \frac{h^2}{a^2}\right )}$
$\displaystyle =\, \frac{-2k}{b^2 \left ( \frac{-k^2}{b^2}\right )}\, =\, \frac{2}{k}$

Find the range of $p$ such that a unique pair of perpendicular tangents can be drawn to the hyperbola $\dfrac{x^2}{(p^2 - 4)} - \dfrac{y^2}{(p^2 + 4p + 3)} = 1$, i.e. the director circle of the given hyperbola is a point.

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

  1. $p > 2$

  2. $p = {-\dfrac{7}{4}}$

  3. $p < -2$

  4. $p = {3}$

Show answer
Correct Option: B
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 a point circle, $a^2 = b^2$ .

$p^2 - 4 = p^2 + 4p + 3$ ---> $4p = -7$ ---> $p = -\dfrac{7}{4}$ . Hence, option (B).

Asymptotes of the hyperbola $xy=4x+3y$ are 

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. x=3, y=4

  2. x=4, y=3

  3. x=2, y=6

  4. x=6, y=2

Show answer
Correct Option: A
Explanation:

correct answer is A.

Given,

$xy=4x+3y$
$\Rightarrow (x-y)=3(y-4)+12$
$\Rightarrow (x-4)(y-4)=12$
$\therefore$ joint equation of asymptotes is $(x-3)(y-4)=0$
hence,
$x-3=0$
$x=3$
and,
$y-4=0$

$y=4$

The angle between the asymptotes to the hyperbola $\dfrac { { x }^{ 2 } }{ 16 } -\dfrac { { y }^{ 2 } }{ 9 } =1$ is

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $\pi -2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $

  2. $\pi -2\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $

  3. $2\tan ^{ -1 }{ \left( \dfrac { 3 }{ 4 } \right) } $

  4. $2\tan ^{ -1 }{ \left( \dfrac { 4 }{ 3 } \right) } $

Show answer
Correct Option: C
Explanation:

Fact: Angle between asymptotes of hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $2\tan^{-1}\left(\dfrac{b}{a}\right)$


Hence angle between the asymptotes to the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ is $2\tan^{-1}\left(\dfrac{3}{4}\right)$

The asymptote of the hyperbole $\dfrac {x^{2}}{a^{2}-y^{2}b^{2}}=1$ from with any tangent to the hyperbola a triangle whose area is $a^{2}tan\lambda$ in magnitude then its eccentricity is ?

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $Sec\lambda$

  2. $csc\lambda$

  3. $sec^{2}\lambda$

  4. $csc^{2}\lambda$

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Correct Option: C

Differential equation of all hyperbolas which pass through the origin, and have their asymptotes parallel to the coordinate axes is?

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $xy\dfrac{d^2y}{dx^2}-2x\left(\dfrac{dy}{dx}\right)^2+2y=0$

  2. $xy\dfrac{d^2y}{dx^2}-2\left(\dfrac{dy}{dx}\right)^2+2y\left(\dfrac{dy}{dx}\right)=0$

  3. $xy\left(\dfrac{d^2y}{dx^2}\right)-2x\left(\dfrac{dy}{dx}\right)^2+2y\dfrac{dy}{dx}=0$

  4. $xy\dfrac{d^2y}{dx^2}+2x\left(\dfrac{dy}{dx}\right)^2+y\left(\dfrac{dy}{dx}\right)=0$

Show answer
Correct Option: A

Area of triangle formed by the tangent at one vertex and asymptotes of the hyperbola xy=2

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. 2sq. units

  2. 3 units

  3. 1 sq. unit

  4. none of these

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Correct Option: A

The product of perpendiculars drawn from any point of a hyperbola with principal axes $2a$ and $2b$ upon its asymptotes is equal to:

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $\frac{a^2b^2}{a^2+b^2}$

  2. $\frac{a^2 +b^2}{a^2b^2}$

  3. $\frac{ab}{a^2+b^2}$

  4. $\frac{ab(a+b)}{\sqrt a+\sqrt b}$

Show answer
Correct Option: A

The angle between the asymptotes of the hyperbola $24x^2 - 8y^2 = 27$ is 

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $90^o$

  2. $60^o$

  3. $120^o$

  4. $45^o$

Show answer
Correct Option: C
Explanation:
$24x^{2}-8y^{2}=27$

divide the above equation with 27 
$\displaystyle \frac{n^{2}}{\dfrac{27}{24}}=\frac{y^{2}}{\dfrac{27}{8}}=1$

$\displaystyle \frac{n^{2}}{\dfrac{9}{8}}-\frac{y^{2}}{\dfrac{27}{8}}=1$

$\displaystyle a^{3}=\frac{9}{8}, b^{2}=\frac{27}{8}$

$\displaystyle a=\frac{3}{2\sqrt{2}},b=\frac{3\sqrt{3}}{2\sqrt{2}}$

let $ 2\alpha $ be the angle between asymptotes 

$2\alpha =2\tan^{-1}\dfrac{b}{a}$

$\displaystyle =2\tan^{-1}\frac{\frac{3\sqrt{3}}{2\sqrt{2}}}{\frac{3}{2\sqrt{2}}}$

$=2\tan^{-1}\sqrt{3}$

$= 2\times \dfrac{\pi}{3}$

$\displaystyle =\frac{2\pi }{3}$  or  $120$
Evaluate the following definite integral:
$\displaystyle \int _{0}^1 \dfrac {1-x}{1+x}dx$
mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $2\log 2+1$

  2. $2\log 2$

  3. $2\log 2-1$

  4. $\log 4$

Show answer
Correct Option: C
Explanation:

$I=\displaystyle \int _{0}^1 \dfrac {1-x}{1+x}dx$ 

$=\displaystyle \int _{0}^1 \dfrac {2-(1+x)}{1+x} dx$ 

$=\displaystyle \int _{0}^1 \left (\dfrac {2}{1+x} -1 \right)dx $

$=\left [2\log (x+1)-x \right] _0^1$

$\Rightarrow \ I=(2\log 2-1)- (2\log 1-0)$ 

$=2\log 2-1$