Tag: two dimensional analytical geometry-ii

Questions Related to two dimensional analytical geometry-ii

If the length of perpendicular drawn from origin to any normal to the ellipse $\cfrac{{x}^{2}}{16}+\cfrac{{y}^{2}}{25}=1$ is $l$, then $l$ cannot be

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

  1. $4$

  2. $5/2$

  3. $1/2$

  4. $2/3$

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Correct Option: A,B,C,D

If the normal at an end of a latus-rectum of an ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through one extremity of the minor axis, the eccentricity of the ellipse is given by:

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

  1. $e^2=5$

  2. $\displaystyle e^2=\frac{\sqrt{5}+1}{2}$

  3. $\displaystyle e=\frac{\sqrt{5}-1}{2}$

  4. $\displaystyle e^2=\frac{\sqrt{5}-1}{2}$

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Correct Option: D
Explanation:

Let $a>b$, then one of latus rectum of the ellipse is $(ae, \cfrac{b^2}{a})$

Thus equation of normal at this point is given by,

$\cfrac{a^2x}{ae}-\cfrac{b^2y}{b^2/a}=a^2e^2$
Given it passes through one of minor axis ,which is $(0,-b)$
$\Rightarrow \cfrac{a^2(0)}{ae}-\cfrac{b^2(-b)}{b^2/a}=a^2e^2$
$\Rightarrow e^2=\cfrac{b}{a}$

Now using $e^2=1-\cfrac{b^2}{a^2}$
we get,  $e^4+e^2-1=0$
$e^2=\cfrac{-1+\sqrt{5}}{2}, \cfrac{-1-\sqrt{5}}{2}$(not possible)

$\therefore e^2=\cfrac{-1+\sqrt{5}}{2}$

If the normal at one end of the latus rectum of an ellipse $\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through one extremity of the minor axis, then:

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

  1. $e^4-e^2+1=0$

  2. $e^2-e+1=0$

  3. $e^2+e+1=0$

  4. $e^4+e^2-1=0$

Show answer
Correct Option: D
Explanation:

Let $a>b$, then one of latus rectum of the ellipse is $(ae, \cfrac{b^2}{a})$
Thus equation of normal at this point is given by,
$\cfrac{a^2x}{ae}-\cfrac{b^2y}{b^2/a}=a^2e^2$
Given it passes through one of minor axis ,which is $(0,-b)$
$\Rightarrow \cfrac{a^2(0)}{ae}+\cfrac{b^2(-b)}{b^2/a}=a^2e^2$
$\Rightarrow e^2=\cfrac{b}{a}$
Now using $e^2=1-\cfrac{b^2}{a^2}$
we get,  $e^4+e^2-1=0$

The normal to the curve x$^2$ = 4y passing (1,2) is 

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

  1. x + y = 3

  2. x - y = 3

  3. x + y = 1

  4. x - y = 1

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Correct Option: A
Explanation:
Given,

$x^2=4y$

$\Rightarrow 2x=4\dfrac{dy}{dx}$

$\therefore \dfrac{dy}{dx}=\dfrac{x}{2}$

$\dfrac{dy}{dx} _{h,k}=\dfrac{h}{2}$

$\dfrac{-1}{\dfrac{dy}{dx} _{h,k}}=-\dfrac{2}{h}$

Equation of normal

$(y-k)=-\dfrac{2}{h}(x-h)$

Given point, $(1,2)$

$(2-k)=-\dfrac{2}{h}(1-h)$

$k=2+\dfrac{2}{h}(1-h)$

$\therefore k=\dfrac{h^2}{4}$

$\Rightarrow \dfrac{h^2}{4}2+\dfrac{2}{h}(1-h)$

upon solving, we get,

$h=2,k=1$

Hence, the equation of normal is 

$(y-1)=\dfrac{-2}{2}(x-2)$

$y-1=-x+2$

$x+y=3$

The line $l x + m y = n$ is a normal to the ellipse $\dfrac { x ^ { 2 } } { a ^ { 2 } } + \dfrac { y ^ { 2 } } { b ^ { 2 } } = 1 ,$ if 

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

  1. $\dfrac { a ^ { 2 } } { l^ { 2 } } + \dfrac { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } { n ^ { 2 } } = \dfrac { b ^ { 2 } } { m ^ { 2 } }$

  2. $\dfrac { a ^ { 2 } } { l ^ { 2 } } + \dfrac { b ^ { 2 } } { m ^ { 2 } } = \dfrac { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } { n ^ { 2 } }$

  3. $\dfrac { \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 } } { n ^ { 2 } } + \dfrac { b ^ { 2 } } { m ^ { 2 } } = \dfrac { a ^ { 2 } } { l ^ { 2 } }$

  4. none of these

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Correct Option: B
Explanation:
Normals of slope m to the ellipse are given by

$y=mx\pm \dfrac{m(a^2-b^2)}{\sqrt{a^2+b^2m^2}}$

so for $y=mx+c$

$c=\pm\dfrac{m(a^2-b^2)}{\sqrt{a^2+b^2m^2}}$

$c^2=\dfrac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$

for $lx+my+n=0$

$y=-\dfrac{1}{m}x-\dfrac{n}{m}$

$c=-n/m$

slope$=-l/m$

$c^2=\dfrac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$

$\dfrac{n^2}{m^2}=\dfrac{\dfrac{l^2}{m^2}\left(a^2-b^2\right)^2}{a^2+b^2\dfrac{l^2}{m^2}}$

$\dfrac{a^2m^2+b^2l^2}{l^2m^2}=\dfrac{(a^2-b^2)^2}{n^2}$

$\dfrac{a^2}{l^2}+\dfrac{b^2}{m^2}=\dfrac{(a^2-b^2)^2}{n^2}$.

The line $5x - 3y = 8\sqrt{2}$ is a normal to the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$. If $\theta$ be the eccentric angle of the foot of this normal , then '$\theta$' is equal to

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

  1. $\dfrac{\pi}{6}$

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

  3. $\dfrac{\pi}{4}$

  4. $\dfrac{\pi}{2}$

Show answer
Correct Option: C
Explanation:
From the equation of ellipse given, we have; $a=5,b=3$

Therefore any point on the ellipse $\left( 5cos\theta , 3sin\theta  \right) .$

Normal at this point to the given ellipse is 
$ 5x sec\theta -3y cosec\theta =25-9=16------(1)$

 Also the equation of given normal is $ 5x-3y=8\sqrt { 2 } -----(2)$

 Also the equation of the normal $(1)$ and $(2)$,

$ \Longrightarrow \dfrac { 5sec\theta  }{ 5 } =\dfrac { 3cosec\theta  }{ 3 } =\dfrac { 16 }{ 8\sqrt { 2 }  }$

$ \Longrightarrow sin\theta =cos\theta =\frac { 1 }{ \sqrt { 2 }  }$

$\Longrightarrow \theta =\frac { \Pi  }{ 4 } $

Option (c) is correct.

Let $L$ be an end of the latus rectum of $y^2 = 4x$. The normal at $L$ meets the curve again at $M$. The normal at $M$ meets the curve again at $N$. The area of $\Delta LMN$ is

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

  1. $\dfrac{1280}{9} sq.$ units

  2. $\dfrac{640}{9} sq.$ units

  3. $\dfrac{320}{9} sq.$ units

  4. $\dfrac{160}{9} sq.$ units

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Correct Option: A

The line $2x+y =3$ cuts the ellipse $4x^2+y^2 =5$ at P and Q . If $\theta$ be the angle between the normals  at these point then $tan \theta$ =

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

  1. $1/2$

  2. $3/4$

  3. $3/5$

  4. $5$

Show answer
Correct Option: C

The equation of the normal to the ellipse $\displaystyle x^{2} + 4y^{2} = 16$ at the end of the latus rectum in the first quadrant is

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

  1. $\displaystyle 2x + \sqrt{3} \left ( y + 3 \right ) = 0$

  2. $\displaystyle 2x = \sqrt{3} \left ( y+ 3 \right )$

  3. $\displaystyle \sqrt{3} x = 2 \left ( y + 3 \right )$

  4. none of these

Show answer
Correct Option: B
Explanation:

Given ellipse may be written as, $\displaystyle \cfrac{x^{2}}{16} + \cfrac{y^{2}}{4} = 1$
$\Rightarrow a^2=16, b^2=4, \therefore e=\sqrt{1-\dfrac14}=\dfrac{\sqrt{3}}2$
Then latus rectum of the ellipse in the first quadrant is $(ae, \cfrac{b^2}{a})\equiv (2\sqrt{3},1)$

Then the point form equation of normal at point $(x _1,y _1)$ is $y-y _1=\displaystyle\frac {y _1a^2}{x _1b^2}(x-x _1), x _1\neq 0$
Thus equation of normal at this point is given by,
$\cfrac{16x}{2\sqrt{3}}-\cfrac{4y}{1}=12$
$\Rightarrow 2x=\sqrt{3}(y+3)$

If the tangent drawn at a point $\left( { t }^{ 2 },2t \right) $ on the parabola ${ y }^{ 2 }=4x$ is same as normal drawn at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ on the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$, then which of following is true.

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

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

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

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

  4. None of these

Show answer
Correct Option: A,B,C
Explanation:

Equation of tangent to ${ y }^{ 2 }=4x$ at $\left( { t }^{ 2 },2t \right) $ is $x=ty-{ t }^{ 2 }$   ....(1)


Equation of normal to ellipse $\displaystyle \frac { { x }^{ 2 } }{ 5 } +\frac { { y }^{ 2 } }{ 4 } =1$ at $\left( \sqrt { 5 } \cos { \alpha  } ,2\sin { \alpha  }  \right) $ is $\sqrt { 5 } \sec { \alpha x-2y\csc { \alpha  } =1 } $    ....(2)

Given (1) $=$ (2)

$\displaystyle \Rightarrow \sqrt { 5 } \sec { \alpha  } =\frac { 2\csc { \alpha  }  }{ t } =-\frac { 1 }{ { t }^{ 2 } } \Rightarrow \cos { \alpha  } =-\sqrt { 5 } { t }^{ 2 }$ and $\sin { \alpha  } =-2t$

$\displaystyle \Rightarrow \cos ^{ 2 }{ \alpha  } +\sin ^{ 2 }{ \alpha  } =5{ t }^{ 4 }+4{ t }^{ 2 }=1\Rightarrow { t }^{ 2 }=\frac { 1 }{ 5 } $   

$\therefore$ (A) is true
and $\displaystyle \frac { \sin { \alpha  }  }{ \cos { \alpha  }  } =-\frac { 2t }{ -\sqrt { 5 } { t }^{ 2 } } =\frac { 2 }{ \sqrt { 5 }  } \times \frac { 1 }{ t } =\frac { 2 }{ \sqrt { 5 }  } \times \left( \pm 5 \right) $

$\therefore \tan { \alpha  } =\pm 2$