Tag: hyperbola

The equation of the normal at $P\left( a\sec { \theta  } ,b\tan {

\theta  }  \right) $ to the hyperbola $\cfrac { { x }^{ 2 } }{ { a

}^{2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is
$ax\cos { \theta  } +by\cot { \theta  } ={ a }^{ 2 }+{ b }^{ 2 }$
This intersects the transverse and conjugate axes at $ L\left( \cfrac { {

a}^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  } ,0 \right) $ and

$M\left(0,\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{  } } \tan {

\theta  }  \right) $ respectively
Let $N(h,k)$. then $h=\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ b } \sec { \theta  } $ and
$k=\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{  } } \tan { \theta  } $
$\Rightarrow

\sec { \theta  } =\cfrac { 2ah }{ { a }^{ 2 }+{ b }^{ 2 } } \quad

,\quad \tan { \theta  } =\cfrac { 2bk }{ { a }^{ 2 }+{ b }^{ 2 } }

\quad$
$\therefore \sec ^{ 2 }{ \theta  } -\tan ^{ 2 }{ \theta  }

=1\quad \Rightarrow 4{ a }^{ 2 }{ h }^{ 2 }-4{ b }^{ 2 }{ k }^{ 2 }={

\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }^{ 2 }$
Thus the locus of

$(h,k)$ is $\Rightarrow 4{ a }^{ 2 }{ x }^{ 2 }-4{ b }^{ 2 }{ y }^{

2}={ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  }^{ 2 }\quad $
Let ${ e } _{ 1 }$ be the eccentricity of this hyperbola. Then
${{

e } _{ 1 } }^{ 2 }=1+\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } =\cfrac { {a

}^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } } =\cfrac { { a }^{ 2 }{ e }^{ 2 }}{ {

a }^{ 2 }({ e }^{ 2 }-1) } $
$\Rightarrow { e } _{ 1 }=\displaystyle \frac { e }{ \sqrt { { e }^{ 2 }-1 }  } $

Given hyperbola is, $\displaystyle \cfrac{x^{2}}{a^2} - \cfrac{y^{2}}{b^2} = 1$
The general equation of normal to hyperbola with slope $m$ is given by,
$y = mx\pm\cfrac{(a^2+b^2)m}{\sqrt{a^2-b^2m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{(a^2+b^2)^2m^2}{a^2-b^2m^2}$
$\Rightarrow (y _1-mx _1)^2(a^2-b^2m^2)=(a^2+b^2)^2m^2$
Clearly, this is a polynomial of degree four so maximum number of normal that can be drawn from point $P$ (any external point) to the hyperbola is $4$ corresponding to four roots of $m$.
Hence, option 'B' is correct.

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 ]$

We know equation of normal to the hyperbola $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$ is given by, $\cfrac{a^2x}{x _1}+\cfrac{b^2y}{y _1}=a^2-b^2$

Thus, required normal is, $5x+\cfrac{16y}{0}=9$
Clearly denominator of second term of L.H.S is $0$ so the equation of line is, $y=0$ 
Hence, option 'A' is correct.  

Required normal is given by,
$\displaystyle \frac{a^2x}{x _1}+\frac{b^2y}{y _1}=a^2+b^2$
$\Rightarrow \displaystyle \frac{16x}{6}+\frac{9y}{(3\sqrt{5}/2)}=25$
$\Rightarrow 8\, \sqrt{5}x\, +\, 18y\, =\, 75\, \sqrt{5}$

Suppose $\displaystyle \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ be a hyperbola and let $\displaystyle \dfrac{x^2}{a^2} - \frac{y^2}{b^2} = - 1$ be its conjugate.
Then their eccentricities are given by $e^2 = \displaystyle \dfrac{a^2 + b^2}{a^2}$ and $\displaystyle e'^2 = \frac{a^2 + b^2}{b^2}$ respectively.
$\therefore \displaystyle \dfrac{1}{e^2} + \dfrac{1}{e'^2} = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$

Equation of normal $\displaystyle Y-y=-\frac { dy }{ dx } \left( X-x \right) $
$\displaystyle \Rightarrow G=\left( x+y\frac { dy }{ dx } ,0 \right) $
According to question
$\displaystyle \left| x+y\frac { dy }{ dx }  \right| =\left| 2x \right| \Rightarrow y\frac { dy }{ dx } =x$ or $\displaystyle y\frac { dy }{ dx } =-3x$
$\Rightarrow ydy=xdx$ or $ydy=-3xdx$
$\displaystyle \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c$ or $\displaystyle \frac { { y }^{ 2 } }{ 2 } =-\frac { 3{ x }^{ 2 } }{ 2 } +c$
$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }=-2c$ or $3{ x }^{ 2 }+{ y }^{ 2 }=2c$