Tag: tangent and normal to a hyperbola

Questions Related to tangent and normal to a hyperbola

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

  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:

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 ]$

Find the equation of normal to the hyperbola $\displaystyle \frac{x^2}{25}\, -\, \displaystyle \frac{y^2}{16}\, =\, 1$ at $(5, 0)$.

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

  1. $y = 0$

  2. $y=-1$

  3. $y=1$

  4. $y=-2$

Show answer
Correct Option: A
Explanation:

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.  

Find the equation of normal to the hyperbola $\displaystyle \frac{x^2}{16}\, -\displaystyle 
\frac{y^2}{9}=1$ at the point $\left ( 6, \displaystyle \frac{3}{2}\sqrt{5}\,\right )$

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

  1. $8\, \sqrt{5}x\, +\, 18y\, =\, 75\, \sqrt{5}$

  2. $4\, \sqrt{5}x\, +\, 9y\, =\, 25\, \sqrt{5}$

  3. $4\, \sqrt{5}x\, +\, 9y\, =\, 75\, \sqrt{5}$

  4. $8\, \sqrt{5}x\, +\, 18y\, =\, 25\, \sqrt{5}$

Show answer
Correct Option: A
Explanation:

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

If e and e' be the eccentricities of a hyperbola and its conjugate, then $\displaystyle \dfrac{1}{e^2} + \dfrac{1}{e'^2} $ is equal to

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

  1. 0

  2. 1

  3. 2

  4. None of these

Show answer
Correct Option: B
Explanation:

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.
$\therefore \displaystyle \dfrac{1}{e^2} + \dfrac{1}{e'^2} = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$

The normal to a curve at $P(x, y)$ meets the x-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is :

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

  1. an ellipse

  2. a parabola

  3. a circle

  4. a hyperbola

Show answer
Correct Option: A,D
Explanation:

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 $3{ x }^{ 2 }+{ y }^{ 2 }=2c$

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 $P$