Tag: normal to a hyperaboal

Assume rectangular hyperbola is $xy = c^2$
Thus equation of tangent and normal at any point 't' are,
$\cfrac{x}{t}+ty=2c$ and $ y-\cfrac{c}{t}=t^2(x-ct)$
Now putting $y=0$ in both the equation we get, $x _1=2ct, x _2=ct-\cfrac{c}{t^3}$
and putting $x=0$ we get, $y _1=\cfrac{2c}{t}, y _2=\cfrac{c}{t}-ct^3$
$\Rightarrow x _1x _2+y _1y _2=2ct(ct-\cfrac{c}{t^3})+\cfrac{2c}{t}(\cfrac{c}{t}-ct^3)=0$
Hence, option 'C' is correct.

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.

The equation of the normal $t _{1}$is  $y-\dfrac{c}{t _{1}}=t _{1}^{2}(x-ct _{1})$

If this passes through $\left ( ct _{2},\dfrac{c}{t _{2}} \right )$

$\dfrac{\mathrm{c}}{\mathrm{t} _{2}}-\dfrac{\mathrm{c}}{\mathrm{t} _{\mathrm{t}}}=\mathrm{t} _{1}^{2}(\mathrm{c}\mathrm{t} _{2}-\mathrm{c}\mathrm{t} _{1})$

$\displaystyle



\Rightarrow-\dfrac{1}{\mathrm{t} _{\mathrm{t}}\mathrm{t} _{2}}=\mathrm{t} _{1}^{2}\Rightarrow

1+\mathrm{t} _{\mathrm{t}}^{3}\mathrm{t} _{2}=0$

$\Rightarrow t _1^3t _2=-1$

Hence, option 'D' is correct.

The equation of the normal at $\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ to the given hyperbola is
$ax\cos { \theta  } +by\cot { \theta  } =\left( { a }^{ 2 }+{ b }^{ 2 } \right) $
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\left{ \cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  }
,0 \right} $
The coordinates of the vertices $A$ and $A'$ are $A(a,0)$ and $A'(-a,0)$ respectively
$\therefore

\quad AG.A'G=\left( -a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec {

\theta  }  \right) \left( a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec

{ \theta  }  \right)$
 $\Rightarrow AG.A'G=\left( -a+a{ e }^{ 2 }\sec { \theta  }  \right) \left( a+a{ e }^{ 2 }\sec { \theta  }  \right) $
$={ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta  } -1 \right) $
Hence, option 'B' is correct.

The equation of the normal at $\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ to the given hyperbola is
$ax\cos { \theta  } +by\cot { \theta  } =\left( { a }^{ 2 }+{ b }^{ 2 } \right) $
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\left{ \cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  }

,0 \right} $
The coordinates of the vertices $A$ and $A'$ are $A(a,0)$ and $A'(-a,0)$ respectively
$\therefore

\quad AG.A'G=\left( -a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec {

\theta  }  \right) \left( a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec

{ \theta  }  \right)$
 $\Rightarrow AG.A'G=\left( -a+a{ e }^{ 2 }\sec { \theta  }  \right) \left( a+a{ e }^{ 2 }\sec { \theta  }  \right) $
$={ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta  } -1 \right) $
Hence, option 'B' is correct.

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.