Tag: two dimensional analytical geometry-ii

Questions Related to two dimensional analytical geometry-ii

The equation of the normal to the ellipse $\displaystyle\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,=\,1$ at the positive end of latus rectum is : 

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

  1. $x\,+\,ey\,+\,e^2a\,=\,0$

  2. $x\,-\,ey\,-\,e^3a\,=\,0$

  3. $x\,-\,ey\,-\,e^2a\,=\,0$

  4. none of these

Show answer
Correct Option: B
Explanation:

$Equation\quad of\quad ellipse:\quad \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\ Co-ordinates\quad of\quad positive\quad latus\quad rectum\quad is:\quad (ae,\frac { { b }^{ 2 } }{ a } )\ Equation\quad of\quad normal\quad at\quad point(x1,y1)\quad is:\ \frac { { a }^{ 2 }x }{ x1 } -\frac { { b }^{ 2 }y }{ y1 } ={ (ae) }^{ 2 }\ \therefore \quad Equation\quad is:\quad \frac { { a }^{ 2 }x }{ ae } -\frac { { b }^{ 2 }y }{ \frac { { b }^{ 2 } }{ a }  } ={ (ae) }^{ 2 }\ Or,\quad \frac { x }{ e } -\frac { y }{ 1 } ={ ae }^{ 2 }\ Or,\quad x-ey-{ ae }^{ 3 }=0$


Option [B]

Area of the triangle formed by the ${x}$ axis, the tangent and normal at $(3,2)$ to the ellipse $\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{8}=1$ is 

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

  1. $5$

  2. $\dfrac{13}{3}$

  3. $\displaystyle \frac{15}{2}$

  4. $\displaystyle \frac{9}{2}$

Show answer
Correct Option: B
Explanation:

Given equation of ellipse $\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{8}=1$
$\displaystyle \frac {dy}{dx}=\displaystyle \frac {-4x}{9y}$
Slope of tangent to ellipse at $(3,2)$ is 
$m=\displaystyle \frac{-2}{3}$

Equation of tangent to ellipse is 
$y-2=-\displaystyle \frac{2}{3}(x-3)$
$\Rightarrow 2x+3y=12$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=6$
So, tangent intersects x-axis at $(6,0)$

Equation of normal to ellipse is 
$y-2=\displaystyle \frac{3}{2}(x-3)$
$\Rightarrow 3x-2y=5$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=\displaystyle \frac{5}{3}$
So, normal intersects x-axis at $\left(\displaystyle \frac{5}{3} ,0\right)$

So, area of triangle $=\displaystyle \frac { 1 }{ 2 } \begin{vmatrix} 3 & 2 & 1 \ 6 & 0 & 1 \ \frac { 5 }{ 3 }  & 0 & 1 \end{vmatrix}$
$=\displaystyle \frac{13}{3}$ sq.units

Find the area of the rectangle formed by the perpendiculars from the center of the ellipse $\displaystyle \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ to the tangent and normal at a point whose eccentric angle is $\displaystyle\frac{\pi}{4}.$ 

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

  1. $\displaystyle \frac { \left( { a }^{ 2 }-{ b }^{ 2 } \right) ab }{ { a }^{ 2 }+{ b }^{ 2 } } $

  2. $\displaystyle \frac { \left( { a }^{ 2 }+{ b }^{ 2 } \right) ab }{ { a }^{ 2 }-{ b }^{ 2 } } $

  3. $\displaystyle \frac { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }{ab( { a }^{ 2 }+{ b }^{ 2 } )} $

  4. $\displaystyle \frac { \left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ab( { a }^{ 2 }-{ b }^{ 2 } )} $

Show answer
Correct Option: A

Assertion (A): Equation of the normal to the ellipse $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ at $P(\displaystyle \frac{\pi}{4})$ is $5x-3y-8\sqrt{2}=0$
Reason (R): Equation of the normal to the ellipse $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $P(x _{1},y _{1})$ is $\displaystyle \frac{a^{2}x}{x _1}-\frac{b^{2}y}{y _1}=a^{2}-b^2$

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

  1. Both A and R are true but R is not the correct explanation of A

  2. Both A and R are true and R is the correct explanation of A

  3. A is true but R is false

  4. A is false but R is True

Show answer
Correct Option: B
Explanation:

Reason is correct.
Equation of normal in parametric form: $\dfrac{ax}{\cos \theta}-\dfrac{by}{\sin \theta}=a^2-b^2$
$\Rightarrow 5\sqrt 2 x-3\sqrt 2 y=25-9=16$
Therefore, assertion is correct but reason is not the correct explanation 

The maximum distance of any normal to the ellipse $\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ from the centre is:

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

  1. $a+b$

  2. $a-b$

  3. $a^{2}+b^{2}$

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

Show answer
Correct Option: B
Explanation:
Equation of any normal to the ellipse is:
$ \cfrac { ax }{ \cos { \theta  } } -\cfrac { by }{ \sin { \theta  } } =ae$ 
Distance from the center is: $ d=\cfrac { ae }{ \sqrt { \cfrac { { a }^{ 2 } }{ { (\cos { \theta  }) }^{ 2 } } +\cfrac { { b }^{ 2 } }{ { (\sin { \theta  }) }^{ 2 } }  }  }$        -------(1) 
$a,e,b$ are fixed for a given ellipse. So, to maximize $d$ we need to minimize the denominator
$ \therefore E={ a }^{ 2 }({ \sec { \theta  }) }^{ 2 }+{ b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 2 }$
$ \cfrac { DE }{ D\theta  } =2{ a }^{ 2 }({ \sec { \theta  }) }^{ 3 }\tan \theta -{ 2b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 3 }\cot \theta =0$
$ \Rightarrow 2{ a }^{ 2 }({ \sec { \theta  }) }^{ 3 }\tan \theta ={ 2b }^{ 2 }{ \left( co\sec { \theta  } \right)  }^{ 3 }\cot \theta $ 
$\Rightarrow { (\tan \theta ) }^{ 4 }=\cfrac { { b }^{ 2 } }{ { a }^{ 2 } } $ 
$\Rightarrow \tan \theta =\sqrt { \cfrac { b }{ a } } $ 
$\therefore  \sin \theta =\sqrt { \cfrac { b }{ a+b }  }$ and 
$\cos \theta =\sqrt { \cfrac { a }{ a+b }  } $ 
Putting these values in equation 1 we get:
$ d=\cfrac { ae }{ \sqrt { \cfrac { { a }^{ 2 }(a+b) }{ a } +\cfrac { { b }^{ 2 }(a+b) }{ b }  }  } $ 
$\Rightarrow d=\cfrac { ae }{ { (a+b) } } $ 
$\Rightarrow d=\cfrac { ae(a-b) }{ { (a+b)(a-b) } } $ 
$\Rightarrow d=a-b$

The maximum distance of the normal to the ellipse $\displaystyle \frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$ from its centre is:

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

  1. $\displaystyle \frac{1}{2}$

  2. $2$

  3. $1$

  4. $4$

Show answer
Correct Option: C
Explanation:
Ellipse : $\cfrac { { x }^{ 2 } }{ 9 } +\cfrac { { y }^{ 2 } }{ 4 } =1$

Equation of the normal,
$\cfrac { { ax }^{  } }{ \cos  { \theta  }  } -\cfrac { { by }^{  } }{ \sin { \theta  } } ={ a }^{ 2 }-{ b }^{ 2 } \\ \therefore \cfrac { { 3x }^{  } }{ \cos  { \theta  }  } -\cfrac { { 2y }^{  } }{ \sin { \theta  } } =5$

Or, $\ { 3x }^{  }\sin { \theta  }-{ 2y }^{  }\cos  { \theta  } =5\cos  { \theta  } \sin { \theta  }$

Distance from origin d= $\cfrac { \left| 0+0-5\cos  { \theta  } \sin { \theta  } \right|  }{ \sqrt { 9{ \left( \cos  { \theta  }  \right)  }^{ 2 }+{ 4\left( \sin { \theta  } \right)  }^{ 2 } }  } $

Or, d=$\cfrac { 5 }{ \sqrt { 9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 } }  } $

To maximize d we need to minimize the denominator.
$E=9{ \left( \csc { \theta  }  \right)  }^{ 2 }+{ 4 }{ \left( \sec { \theta  }  \right)  }^{ 2 }\ then,\quad \\\cfrac { dE }{ d\theta  } =-18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \ For\quad \\Minimizing,\quad \cfrac { dE }{ d\theta  } =0\\ \therefore -18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } +8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } =0 \\Or,18{ \left( \csc { \theta  }  \right)  }^{ 2 }\cot { \theta  } =8{ \left( \sec { \theta  }  \right)  }^{ 2 }\tan { \theta  } \\ Or,{ \left( \tan { \theta  }  \right)  }^{ 4 }=\cfrac { 9 }{ 4 } \\ Or,\quad \tan { \theta  } =\sqrt { \cfrac { 3 }{ 2 }  } \\ \therefore \csc { \theta  } =\sqrt { \cfrac { 5 }{ 3 }  } \quad \quad and\quad \quad \sec { \theta  } =\sqrt { \cfrac { 5 }{ 2 }  } $

 On putting the values in d we get,
$d=\cfrac { 5 }{ \sqrt { 15+10 }  } \\ Or,\quad d=1$

lf the tangent drawn at a point $(t^{2},2t)$ on the parabola $y^{2}=4x$ is same as normal drawn at $(\sqrt{5}\cos\alpha, 2\sin\alpha)$ on the ellipse $\displaystyle \frac{x^{2}}{5}+\frac{y^{2}}{4}=1$, then which of following is not true?  

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

  1. $t=\displaystyle \pm\frac{1}{\sqrt{5}}$

  2. $\alpha=-\tan^{-1}2$

  3. $\alpha=\tan^{-1}2$

  4. $\alpha=\tan^{-1}4$

Show answer
Correct Option: D

If the line $x\cos { \alpha  } +y\sin { \alpha  } =p$ be normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, then

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

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

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

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

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

Show answer
Correct Option: D
Explanation:

The equation of any normal to $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is 
       $ax\sec { \phi  } -by\csc { \phi  } ={ a }^{ 2 }-{ b }^{ 2 }$              ......(i)
The straight line $x\cos { \alpha  } +y\sin { \alpha  } =p$ will be a normal to the ellipse $\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$, if equation (i) and $x\cos { \alpha  } +y\sin { \alpha  } =p$ represent the same line.
$\therefore \dfrac { a\sec { \phi  }  }{ \cos { \alpha  }  } =\dfrac { -b\csc { \phi  }  }{ \sin { \alpha  }  } =\dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ p } $
$\Rightarrow \cos { \phi  } =\dfrac { ap }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \cos { \alpha  }  } $
$\sin { \phi  } =\dfrac { -bp }{ \left( { a }^{ 2 }-{ b }^{ 2 } \right) \sin { \alpha  }  } $
$\because \sin ^{ 2 }{ \phi  } +\cos ^{ 2 }{ \phi  } =1$
$\Rightarrow \dfrac { { b }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\sin ^{ 2 }{ \alpha  }  } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }\cos ^{ 2 }{ \alpha  }  } =1$
$\Rightarrow { p }^{ 2 }\left( { b }^{ 2 }\csc ^{ 2 }{ \alpha  } +{ a }^{ 2 }\sec ^{ 2 }{ \alpha  }  \right) ={ \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 }$

If the line $x \cos a + y \sin a = p$ be normal to the ellipse $\dfrac{x^2}{a^2}$ $+\dfrac{y^2}{b^2}$ = 1 then

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

  1. $p^2(a^2\cos^2a+b^2\sin^2a)=a^2-b^2$

  2. $p^2(a^2\cos^2a+b^2\sin^2a)=(a^2-b^2)^2$

  3. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)$

  4. $p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$

Show answer
Correct Option: A
Explanation:

A line $y=mx+c$ is normal to ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if $c^2=m^2\dfrac{(a^2-b^2)^2}{a^2+m^2b^2}$
Given equation $x\cos a+y\sin a=p\Rightarrow y=-(\dfrac{\cos a}{\sin a})x+\dfrac{p}{\sin a}$
Here $m=-\cot a, c=\dfrac{p}{\sin a}$
Substituting in the formulae, we get
$\dfrac{p^2}{\sin ^2a}=\dfrac{\cos ^2a}{sin^2a}\times \dfrac{(a^2-b^2)^2}{a^2+(\cot^2a) b^2}$
After simplification, we get
$p^2\dfrac{(a^2\sin^2a+b^2\cos^2a)}{\sin^2a\cos^2a}=(a^2-b^2)^2$
$p^2(a^2\sec^2a+b^2\csc^2a)=(a^2-b^2)^2$

If the normal at the point $P(\theta)$ to the ellipse $\dfrac {x^{2}}{14} + \dfrac {y^{2}}{5} = 1$ intersects it again at the point $Q(2\theta)$, then $\cos \theta$ is equal to

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

  1. $2/3$

  2. $-2/3$

  3. $3/4$

  4. None of these

Show answer
Correct Option: B
Explanation:
Normal at the point P(theta) to the ellipse $\dfrac{x²}{14} + \dfrac{y²}{5 }= 1$ intersects it again at the point $Q(2\  \theta). $

we know, standard equation of ellipse is 

$\dfrac{x²}{a²} + \dfrac{y²}{b²} = 1 $ compare it with given equation

so, $ a² = 14$ then, $a = √14 $

$b² = 5$ then, $b = √5 $

now equation of normal passing through point $P(\theta)$ is given by, 

$\dfrac{ax}{cos \theta} - \dfrac{by}{sin \theta} = a² - b². $

or, $\dfrac{\sqrt14x}{cos \theta} -\dfrac{ \sqrt5y}{sin \theta} = 14 - 5 = 9$ ....(1) 

it again meets the curve at the point $Q(2\theta) $

so, $Q(2\theta) = (√14cos2\theta, √5sin2\theta) $

now, put it in equation (1), 

or, $\dfrac{14cos2\theta}{\cos \theta} - \dfrac{5sin2\theta}{\sin\theta} = 9$ 

or, ${14(2cos² \theta - 1)}{\cos \theta} - \dfrac{10sin\theta cos \theta}{\sin \theta} = 9$

or, $28cos \theta - 14sec \theta - 10cos \theta = 9$

or, $18cos \theta - \dfrac{14}{cos \theta} = 9$

or, $18cos²\theta - 14 - 9cos \theta = 0$

or, $18cos²\theta -21cos \theta + 12cos\theta - 14 = 0$

or, $3cos \theta(6cos \theta - 7) + 2(cos \theta - 7) = 0$

or, $(3cos \theta + 2)(6cos \theta - 7) = 0$

or, $cos \theta = \dfrac{-2}{3} $