Tag: two dimensional analytical geometry-ii

Questions Related to two dimensional analytical geometry-ii

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} $

If a normal is drawn at point $P$ of ellipse $ \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, then the maximum distance from centre of ellipse will be $a-b$ 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. True

  2. False

Show answer
Correct Option: A

If the normal at any point $P$ of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ meets the axes in $G$ and $g$ respectively, then $|PG| : |Pg|$ is equal to 

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $a:b$

  2. $a^2:b^2$

  3. $b^2:a^2$

  4. $b:a$

Show answer
Correct Option: A

One foot of normal of the ellipse $4x^2$ $+$ 9$y^2$ $= 36 $, that is parallel to the line $2x + y = 3 $, is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $\left ( \dfrac{9}{8}, \dfrac{5}{8} \right )$

  2. $\left ( \dfrac{9}{8}, \dfrac{8}{5} \right )$

  3. $\left ( \dfrac{8}{9}, \dfrac{8}{5} \right )$

  4. None

Show answer
Correct Option: B

If the normal at any point on the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ meets the axes in $G$ and $g$ respectively, then $PG:Pg=$

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $a:b$

  2. ${ a }^{ 2 }:{ b }^{ 2 }$

  3. $b:a$

  4. ${ b }^{ 2 }:{ a }^{ 2 }$

Show answer
Correct Option: A

The equation of normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

  1. $y-  3 = 0$

  2. $y + 3 = 0$

  3. $x$-axis

  4. $y$-axis

Show answer
Correct Option: C
Explanation:

$9x^2+5y^2=45$


Differentiating with respect to $x$, we get,

$18x+10y\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=-\dfrac{18x}{10y}$

Therefore, Slope of tangent at $(0,3)=\dfrac{-0}{30}=0$

Slope of normal = $\dfrac{1}{0}$

The equation of normal is :
$y-3=\dfrac{1}{0}(x-0)$
$x-0=0$
$x=0$