Tag: maths

Questions Related to maths

Line x cos$\alpha $+yin$\alpha $=p is a normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $, if

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

  1. $a^{2}sec^{2}\alpha -b^{2}cosec^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$

  2. $a^{2}sec^{2}\alpha+b^{2}cosec^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$

  3. $a^{2}cos^{2}\alpha -b^{2}sin^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$

  4. $a^{2}cos^{2}\alpha+b^{2}sin^{2}\alpha =\frac{(a^{2}+b^{2})^{2}}{p^{^{2}}}$

Show answer
Correct Option: A

Line $ x \cos \alpha + y \sin \alpha = p $ is a normal to the hyperbola $ \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 $, if 

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

  1. $ a ^ { 2 } \sec ^ { 2 } \alpha - b ^ { 2 } \csc ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $

  2. $ a ^ { 2 } \sec ^ { 2 } x + b ^ { 2 } \csc ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $

  3. $ a ^ { 2 } \cos ^ { 2 } \alpha - b ^ { 2 } \sin ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $

  4. $ a ^ { 2 } \cos ^ { 2 } \alpha + b ^ { 2 } \sin ^ { 2 } \alpha = \frac { \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } } { p ^ { 2 } } $

Show answer
Correct Option: A

A straight line is drawn parallel to the conjugate axis of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ to meet it and the conjugate hyperbola respectively in the point $P$ and $Q$. The normals at $p$ and $Q$ to the curves meet on 

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

  1. $x-axis$

  2. $y-axis$

  3. $y=x$

  4. $y=-x$

Show answer
Correct Option: A
Explanation:
Equation of hyperbola $\to \dfrac {x^2}{9^2}-\dfrac {y^2}{6^2}=1$
Conjugate hyperbola $\to \dfrac {y^2}{6^2}-\dfrac {x^2}{9^2}=1$
Line $||$ to conjugate axis of hyperbola $(y-axis)$ is drawn to meet conjugate axis at $P$ & $Q$ which are symmetric points about $x-$axis then
$P(a\tan \theta , b\sin theta)$
$Q(a \tan \theta , -b\sin \theta)$
Normal at $P\to \dfrac {ax}{\tan \theta} +\dfrac {6y}{\sec \theta}=a^2+b^2$
Normal at $Q\to \dfrac {ax}{\tan \theta}-\dfrac {6y}{\sec \theta}=a^2+b^2$
Let us find out interrection
$\dfrac {b}{\sec \theta}=\dfrac {-6y}{\sec \theta} $
$y=0$
$9+$ lies on $x-$axis
$A$ is correct


If the normal at $\left (ct _1,\dfrac { c}{t _1}\right)$ on the hyperbola $xy = c^2$ cuts the hyperbola again at $\left (ct _2, \dfrac {c}{t _2}\right)$, then $t _2^3 t _2$ $=$ 

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

  1. $2$

  2. $-2$

  3. $-1$

  4. $1$

Show answer
Correct Option: C
Explanation:

The equation of hyperbola is $xy=c^2$ and point $(ct _1,\dfrac{c}{t _1})$ lies on it.


Let us find the equation of the normal.


Equation of hyprbola can be written as $y=\dfrac{c^2}{x}$ and therefore slope of tangent is given by first derivative i.e. $\dfrac{dy}{dx}=−\dfrac{c^2}{x^2}$


hence slope of normal is given by $\dfrac{x^2}{c^2}$ and at $(ct _1,\dfrac{c}{t _1})$ is $t^2$ and its equation is


$y=t _1^2(x−ct _1)+\dfrac{c}{t _1}$


or $xt _1^3−yt _1−ct _1^4+c=0$


As this passes through $(ct _2,\dfrac{c}{t _2})$


$ct _2t _1^3−\dfrac{c}{t _2}t _1−ct _1^4+c=0$


or $ct _1^3(t _2−t _1)+\dfrac{c}{t _2}(t _2−t _1)=0$


as $t _1\neq t _2, t _1−t _2\neq 0$ and dividing by it we get


$ct _2^3=−\dfrac{c}{t _2}$


Or $ t _2^3t _2=−1$

If the tangent and normal to a rectangular hyperbola cut off intercepts $x _1$ and $x _2$ on one axis and $y _1$ and $y _2$ on the other axis, then

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

  1. $x _1y _1+x _2y _2=0$

  2. $x _1y _2+x _2y _1=0$

  3. $x _1x _2+y _1y _2=0$

  4. none of these

Show answer
Correct Option: C
Explanation:

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.

The number of normal to the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ from an external point is

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

  1. $2$

  2. $4$

  3. $6$

  4. $5$

Show answer
Correct Option: B
Explanation:

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 normal to the rectangular hyperbola $xy=-c^2$ at the point $'t _1'$ meets the curve again at the point $'t _2'$. The value of $t _1^3 \cdot t _2$ is

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

  1. $1$

  2. $c$

  3. $-c$

  4. $-1$

Show answer
Correct Option: D
Explanation:

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.

If the normal at '$\theta $' on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ meets the transverse axis at $G$  and $A$ and $A'$ are the vertices of the hyperbola, then $AG.A'G$ $=$

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

  1. ${ a }^{ 2 }\left( { e }^{ 2 }\sec ^{ 2 }{ \theta } -1 \right) $

  2. ${ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta } -1 \right) $

  3. ${ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta } +1 \right) $

  4. none of these

Show answer
Correct Option: B
Explanation:

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.

If the normal at $'\theta'$ on the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ meets the transverse axis at G, and A and A' are the vertices of the hyperbola, then AG.A'G $=$

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

  1. $a^2 (e^2 sec^2 \theta -1)$

  2. $a^2 (e^4 sec^2 \theta - 1)$

  3. $a^2 (e^4 sec^2 \theta + 1)$

  4. none of these

Show answer
Correct Option: B
Explanation:

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 normal at $\left( at,\dfrac { a }{ t }  \right)$ to the hyperbola $xy={ a }^{ 2 }$ is ________________________.

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

  1. ${ xt }^{ 3 }-yt+{ at }^{ 4 }-a=0$

  2. ${ xt }^{ 3 }-yt-{ at }^{ 4 }+a=0$

  3. ${ xt }^{ 3 }+yt+{ at }^{ 4 }-a=0$

  4. ${ xt }^{ 3 }+yt-{ at }^{ 4 }-a=0$

Show answer
Correct Option: B