Tag: hyperbola

Questions Related to hyperbola

lf the line $ax+by+c=0$ is a normal to the curve $xy=1$, then  :

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $a>0,b>0$

  2. $a>0,b<0$

  3. $a<0,b>0$

  4. $a<0,b<0$

Show answer
Correct Option: B,C
Explanation:

$xt^{3}-yt-t^{4}+120$
$eq^{n}$ of the normal at $t$ will same as the line $ax+by+c=0$
$\therefore \dfrac{t^{3}}{a}=\dfrac{-t}{b}=\dfrac{-1-t^{4}}{c}$
$t^{2}=\dfrac{-a}{b}$
$\therefore ab<0$

The equation of the normal at the positive end of the latusrectum of the hyperbola $x^2-3y^2=144$ is

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $\sqrt{3}x+2y=32$

  2. $\sqrt{3}x-3y=48$

  3. $3x+\sqrt{3}y=48$

  4. $3x-\sqrt{3}y=48$

Show answer
Correct Option: A
Explanation:
The given hyperbola has the equation $\dfrac{x^2}{12^2}-\dfrac{y^2}{(4\sqrt{3})^2}=1$
Eccentricity of hyperbola = $e = \dfrac{\sqrt{a^2+b^2}}{a}=\dfrac{2}{\sqrt{3}}$
Now, equation of positive latus rectum is $x=ae=8\sqrt{3}$
The end-points of latus rectum are calculated as
$\dfrac{(8\sqrt{3})^2}{12^2}-\dfrac{y^2}{(4\sqrt{3})^2}=1$
$\therefore \dfrac{16}{12}-\dfrac{y^2}{48}=1$
$\therefore y^2=\dfrac{48}{3}$
$\therefore y=\pm 4$
Hence, the positive end is $(8\sqrt{3},4)$.
Now, equation of normal at any point $(x _1,y _1)$ is $\dfrac{a^2x}{x _1}+\dfrac{b^2y}{y _1}=a^2e^2$
$\therefore \dfrac{144x}{8\sqrt{3}}+\dfrac{48y}{4}=48\times 4$
$\therefore 6\sqrt{3}x+12y=48\times 4$
$\therefore \sqrt{3}x+2y=32$
This is the required answer.

Which one of the following points does not lie on the normal to the hyperbola, $\cfrac { { x }^{ 2 } }{ 16 } -\cfrac { { y }^{ 2 } }{ 9 } =1$ drawn at the point $\left( 8,3\sqrt { 3 }  \right) $?

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $\left( 13,-\cfrac { 1 }{ \sqrt { 3 } } \right) $

  2. $\left( 12,\cfrac { 1 }{ \sqrt { 3 } } \right) $

  3. $\left( 11,\sqrt { 3 } \right) $

  4. $\left( 10,\sqrt { 3 } \right) $

Show answer
Correct Option: D
Explanation:

$\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$   $\Rightarrow \dfrac{2x}{16} - \dfrac{2y}{9} \dfrac{dy}{dx} = 0$

At $(8, 3\sqrt{3})$
$\dfrac{2\times 8}{16} - \dfrac{2\times 3\sqrt{3}}{9}\dfrac{dy}{dx}=0 \Rightarrow \dfrac{3}{2\sqrt{3}}=\dfrac{dy}{dx}$

Therefore slope of normal $=-\dfrac{1}{\frac{dy}{dx}}$
Equation of normal at $(8, 3\sqrt{3})$, 
$y-3\sqrt{3} = -\dfrac{2}{\sqrt{3}}(x-8)$
Clearly, option (D) does not lies on it.

Let $A\left( A\sec { \theta  } ,3\tan { \theta  }  \right) $ and $B\left( A\sec { \phi  } ,3\tan { \phi  }  \right) $ where $\theta +\phi =\cfrac { \pi  }{ 2 } $, be two points on the hyperbola $\cfrac { { x }^{ 2 } }{ 4 } -\cfrac { { y }^{ 2 } }{ 9 } =1$. If $\left( \alpha ,\beta  \right) $ is the point of intersection of normals to the hyperbola at $A$ and $B$, then $\beta=$

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $\cfrac { -13 }{ 3 } $

  2. $\cfrac { 13 }{ 3 } $

  3. $\cfrac { 3 }{ 13 } $

  4. $\cfrac { -3 }{ 13 } $

Show answer
Correct Option: A
Explanation:

equation of hyperbola at point $A(2\sec{\theta} , 3\tan{\theta})$ is

$y+\dfrac{2}{3}\sin{\theta}x = \dfrac{13}{3}\tan{\theta}$   -------  $(i)$

and at point $B(2sec{\phi} , 3\tan{\phi})$ is
$y+\dfrac{2}{3}\sin{\phi}x = \dfrac{13}{3}\tan{\phi}$
now 

putting $\phi = \dfrac{\pi}{2} - \theta$


$y+\dfrac{2}{3}\cos{\theta}x = \dfrac{13}{3}\cot{\theta}$   -----  $(ii)$

now multiplying eq.(i) with  $\cos{\theta}$   and eq (ii) with  $\sin{\theta}$

then subtract both equation we find value of $\beta = -\dfrac{13}{3}$

If the sum of the slopes of the normal from a point P to the hyperbola $xy = {c^2}$is equal to $\lambda (\lambda  \in {R^ + })$,then the locus of point P is 

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. ${x^2} = \lambda {c^2}$

  2. ${y^2} = \lambda {c^2}$

  3. ${xy} = \lambda {c^2}$

  4. ${y^2} = {c^2}$

Show answer
Correct Option: A
Explanation:
Equation of rectangular hyperbola is $xy={c}^{2}$

Its rectangular coordinates are $\left(ct,\dfrac{c}{t}\right)$

Equation of normal is $c{t}^{4}-x{t}^{3}+ty-c=0$

Slope$=\dfrac{-coefficient\,of\,x}{coefficient\,of\,y}=\dfrac{{t}^{3}}{t}={t}^{2}$

The normal passes through the point $P\left(h,k\right)$

$\Rightarrow\,c{t}^{4}-h{t}^{3}+tk-c=0$

$\therefore\,$ there exists $4$ roots ${t} _{1},{t} _{2},{t} _{3}$ and ${t} _{4}$

Sum of the roots$={t} _{1}+{t} _{2}+{t} _{3}+{t} _{4}=\dfrac{-coefficient\,of\,{t}^{3}}{coefficient\,of\,{t}^{4}}=\dfrac{-\left(-h\right)}{c}=\dfrac{h}{c}$

Sum of the roots taken two at a time$=\sum{{t} _{i}{t} _{j}}={t} _{1}{t} _{2}+{t} _{2}{t} _{3}+{t} _{3}{t} _{4}+{t} _{4}{t} _{1}+{t} _{2}{t} _{4}+{t} _{1}{t} _{3}=\dfrac{-coefficient\,of\,{t}^{2}}{coefficient\,of\,{t}^{4}}=\dfrac{0}{c}=0$

Now,$\sum{{{t} _{i}}^{2}}=\sum{{\left({t} _{i}\right)}^{2}}$ using ${a}^{2}+{b}^{2}={\left(a+b\right)}^{2}$ for $ab=0$

Sum of squares of slopes of normal from $P$ is
${{t} _{1}}^{2}+{{t} _{2}}^{2}+{{t} _{3}}^{2}+{{t} _{4}}^{2}={\left({t} _{1}+{t} _{2}+{t} _{3}+{t} _{4}\right)}^{2}$

$\Rightarrow\,\lambda={\left(\dfrac{h}{c}\right)}^{2}$

$\Rightarrow\,{h}^{2}=\lambda{c}^{2}$

Replace $h\rightarrow\,x$ we get

${x}^{2}=\lambda{c}^{2}$

$\therefore\,{x}^{2}=\lambda{c}^{2}$ is the required locus at $P$

Let $P\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ and $Q\left( a\sec { \phi  } ,b\tan { \phi  }  \right) $, where $\theta +\phi =\dfrac {\pi}{2} $, be the two points on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$. If $(h,k)$ is the point of intersection of the normals of $P$ and $Q$, then $k$ is equal to

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

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

  2. $-\left[\dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a }\right] $

  3. $\dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b } } $

  4. $-\left[\dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ b } \right]$

Show answer
Correct Option: D
Explanation:

The equations of the normal at P is $ax+bycosec\theta =\left ( a^{2}+b^{2} \right )\sec \theta $          (i)

and the equation of the normal at $Q\left ( a\sec \phi , b\sec \phi  \right )$ is
$ax+by cosec\phi =\left ( a^{2}+b^{2} \right )\sec \phi $          (ii)
Subtracting (ii) from (i) we get

   $\displaystyle y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec \phi }{cosec \theta -cosec \phi }$

So $\displaystyle k=y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec

\left ( \pi /2-\theta  \right )}{cosec \theta -cosec \left ( \pi

/2-\theta  \right )}$          $\left [ \because \theta +\phi =\pi /2

\right ]$
    
$\displaystyle =\frac{a^{2}+b^{2}}{b}.\frac{\sec

\theta -cosec \theta }{cosec \theta -\sec \theta }=-\left [

\frac{a^{2}+b^{2}}{b} \right ]$
Hence, option 'D' is correct.

If a normal of slope $m$ to the parabola ${ y }^{ 2 }=4ax$ touches the hyperbola ${ x }^{ 2 }-{ y }^{ 2 }={ a^2 }$, then

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. ${ m }^{ 6 }-{ 4m }^{ 4 }-{ 3m }^{ 2 }+1=0$

  2. ${ m }^{ 6 }-{ 4m }^{ 4 }+{ 3m }^{ 2 }-1=0$

  3. ${ m }^{ 6 }+{ 4m }^{ 4 }-{ 3m }^{ 2 }+1=0$

  4. ${ m }^{ 6 }+{ 4m }^{ 4 }+{ 3m }^{ 2 }+1=0$

Show answer
Correct Option: D
Explanation:

Equation of normal with slope $'m'$ to the parabola $y^2=4ax$ is given by,
$y=mx-2am-am^3$
Also this line touches the hyperbola $x^2-y^2=a^2$
thus using condition of tangency to the hyperbola, $c^2=a^2m^2-b^2$
$(-2am-am^3)^2=a^2(m^2-1)$
$\Rightarrow 4m^2+m^6+4m^4=m^2-1$
$\Rightarrow m^6+4m^4+3m^2+1=0$
Hence, option 'D' is correct.

If a normal of slope $m$ to the parabola $y^2 = 4ax$ touches the hyperbola $x^2 - y^2 = a^2$, then

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $m^6 - 4m^4 - 3m^2 + 1 =0$

  2. $m^6 - 4m^4 + 3m^2 - 1 = 0$

  3. $m^6 + 4m^4 - 3m^2 + 1 = 0$

  4. $m^6 + 4m^4 + 3m^2 + 1 = 0$

Show answer
Correct Option: D
Explanation:

Equation of normal with slope $'m'$ to the parabola $y^2=4ax$ is given by,
$y = mx-2am-am^3$ (i)

Now given (i) is tangent to the hyperbola $x^2-y^2=a^2$

Thus using condition of tangency, $c^2= a^2m^2-a^2$

$\Rightarrow (2am+am^3)^2=a^2(m^2-1)$

$\Rightarrow (2m+m^3)^2=m^2-1\Rightarrow m^6+4m^4+3m^2+1=0$

Let P $(asec \theta,\, btan \theta)$ and Q $(asec \phi,\, btan \phi)$, where $\theta\, +\, \phi\, =\, \displaystyle \frac{\pi}{2}$, be two points on the hyperbola $\displaystyle \frac{x^2}{a^2}\, -\, \frac{y^2}{b^2}\, =\, 1$. If (h, k) is the point of intersection of the normals at P & Q, then k is equal to

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

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

  2. $\displaystyle - \left (\frac{a^2\, +\, b^2}{a}\right )$

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

  4. $\displaystyle - \left (\frac{a^2\, +\, b^2}{b}\right )$

Show answer
Correct Option: D
Explanation:

Normal at $\theta,\, \phi$ are
$\displaystyle \left {

\begin{matrix} ax\, cos\, \theta & + & by\, cot\, \theta\, =\,

a^2\, +\, b^2 \ ax\, cos\, \phi & + & by\, cot\, \phi\, =\,

a^2\, +\, b^2 \end{matrix}\right.$
where $\displaystyle \phi\, =\, \frac{\pi}{2}\, -\, \theta$ and these passes through (h, k).

$\therefore\, ah\, cos \theta\, +\, bk\, cot \theta\, =\, a^2\, +\, b^2$ .....(i)
$ah\, sin \theta\, +\, bk\, tan \theta\, =\, a^2\, +\, b^2$ .....(ii)
Multiply (i) by $sin \theta$ & (ii) by $cos \theta$ & subtract them, 
we get
$\Rightarrow\, (bk\, +\, a^2\, +\, b^2)\, (sin \theta\, -\, cos \theta)\, =\, 0$
$k =-(\cfrac{a^2 + b^2}{b})$
Hence, option 'D' is correct.

From any point R two normals which are right angled to one another are drawn to the hyperbola $\displaystyle \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\left ( a>b \right )$ If the feet of the normals are P and Q then the locus of the circumcentre of the triangle PQR is

normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

  1. $\displaystyle \frac{x^{2}+y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \right )^{2}$

  2. $\displaystyle \frac{x^{2}-y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2}$

  3. $\displaystyle \frac{x^{2}+y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2}$

  4. $\displaystyle \frac{x^{2}+y^{2}}{a^{2}+b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2}$

Show answer
Correct Option: C
Explanation:

Clearly tangent at P and Q intersect at right-angles at S (say)
$ \displaystyle \Rightarrow $ PSQR is cyclic


$ \displaystyle \Rightarrow $ S lies on director circle of hyperbola
$ \displaystyle \Rightarrow S=\sqrt{a^{2}-b^{2}}\cos \theta , \sqrt{a^{2}-b^{2}}\sin \theta  $

$ \displaystyle \Rightarrow   $ Chord with middle point (h,k) i.e. circumcentre will be same as equation of chord of contact w.r.$ \displaystyle \Rightarrow \perp  $ s

$ \displaystyle \Rightarrow \frac{xh}{a^{2}}-\frac{yk}{b^{2}}=\frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}$ and $\frac{x\sqrt{a^{2}-b^{2}\cos \theta }}{a^{2}}-\frac{y\sqrt{a^{2}-b^{2}}\cos\theta }{b^{2}}=1 $ are identical comparing and solving we get locus as $ \displaystyle \frac{x^{2}+y^{2}}{a^{2}-b^{2}}=\left ( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} \right )^{2} $