Tag: maths

Questions Related to maths

Number of distinct normal lines that can be drawn to the ellipse $\displaystyle \frac{x^2}{169} + \frac{y^2}{25} = 1$ from the point $P(0, 6)$ is:

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

  1. One

  2. Two

  3. Three

  4. Four

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Correct Option: C

If the normal at any point $P$ on the ellipse $\displaystyle\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ 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. $b^2:a^2$

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Correct Option: D
Explanation:

Let $P\equiv (a\cos\theta, b\sin\theta)$
Thus equation of normal to the given ellipse at 'P' is given by,
$ax\sec\theta-by cosec\theta=a^2-b^2$
$\therefore G \equiv ((a-\cfrac{b^2}{a})\cos\theta,0), g\equiv (0,(b-\cfrac{a^2}{b})\sin\theta)$
Thus $PG = \sqrt{\cfrac{b^4}{a^2}\cos^2\theta+b^2\sin^2\theta}=\cfrac{b}{a}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
and $Pg = \sqrt{a^2\cos^2\theta+\cfrac{a^4}{b^2}\sin^2\theta}=\cfrac{a}{b}\sqrt{b^2\cos^2\theta+a^2\sin^2\theta}$
$\therefore PG:Pg = \cfrac{b^2}{a^2} $

The eccentricity of an ellipse whose centre is  at the origin is $\dfrac{1}{2}.$ If one of its directrices is $x =  - 4,$ then the equation of the normal to it at $\left( {1,\dfrac{3}{2}} \right)$ is

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

  1. $2y-x=2$

  2. $4x-2y=1$

  3. $4x+2y=7$

  4. $x+2y=4$

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Correct Option: B
Explanation:
Given: Eccentricity of ellipse$=\dfrac{1}{2}$
Now, $\dfrac{a}{e}=-4$
$\Rightarrow a=4\times\dfrac{1}{2}=2$
$\therefore {b}^{2}={a}^{2}\left(1-{e}^{2}\right)$
$\Rightarrow {a}^{2}\left(1-\dfrac{1}{4}\right)=3$
$\Rightarrow \dfrac{3}{4}{a}^{2}=3$
$\therefore {a}^{2}=4$
$\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{3}=1$
Differentiating w.r.t $x$ we get
$\dfrac{2x}{4}+\dfrac{2y}{3}\dfrac{dy}{dx}=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{-2x}{4}}{\dfrac{2y}{3}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-3x}{4y}$
$\Rightarrow \left[\dfrac{dy}{dx}\right] _{\left(1,\frac{3}{2}\right)}=\dfrac{-3}{4}\times\dfrac{2}{3}=\dfrac{-1}{2}$
Equation of normal at $\left(1,\dfrac{3}{2}\right)$ is 
$y-\dfrac{3}{2}=2\left(x-1\right)$
$\Rightarrow 2y-3=4x-4$
$\Rightarrow 4x-2y=1$ is the equation of the normal.

Tangents are drawn to the ellipse $ \displaystyle \frac{x^2}{a^2}+\displaystyle \frac{y^2}{b^2}=1 $ at points where it is intersected by the line $ \ell x+my+n=0 $. Find the point of intersection of tangents at these points.

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

  1. $ \displaystyle \frac{-a^2}{n},\displaystyle \frac{-b^2m}{n} $

  2. $ \displaystyle \frac{-a^2\ell}{n},\displaystyle \frac{-b^2m}{n} $

  3. $ \displaystyle \frac{-a^2\ell}{n},\displaystyle \frac{-b^2}{n} $

  4. None of these

Show answer
Correct Option: B
Explanation:

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the point of intersection of the
Line $lx+my+n=0$ and the ellipse $\cfrac { { x }^{ 2 }

}{ {a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of tangent at $P$ is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } +\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the point of intersection of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { -{ b }^{ 2 }m }{ n } $
Therefore, the point of intersection of given line and  the given ellipse is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $
Hence, option 'B' is correct.

A ray emanating from the point $(4, 0)$ is incident on the ellipse $9x^2\, +\, 25y^2\, =\, 225$ at the point $P$ with abscissa $3$. Find the equation of the reflected ray after first reflection.

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

  1. $12x + 5y = 48$

  2. $12x - 5y = 48$

  3. $-12x - 35y = 48$

  4. $-12x + 35y = 48$

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Correct Option: C,D
Explanation:

Solution:

Given equation of ellipse: $9x^2+25y^2=225$
$\cfrac{x^2}{25}+\cfrac{y^2}{9}=1$
Let $P(3,y _1)$
or, $9\times3^2+25y _1^2=225$
or, $25y _1^2=144$
or, $y _1=\pm\cfrac{12}{5}$
$\cfrac{x^2}{25}+\cfrac{y^2}{9}=1$
$\cfrac{x^2}{5^2}+\cfrac{y^2}{3^2}=1$
or, $e^2=1-\cfrac{b^2}{a^2}$

or, $e^2=1-\cfrac{9}{25}=\cfrac{16}{25}$
or, $e=\cfrac{4}{5}$
$ae=5\times\cfrac45=4$
It means the given point $(4,0)$ is focus of the ellipse.
We know that rays emanating from the one focus passes through other focus i.e $(-4,0)$
So, Equation of reflected ray $\Rightarrow y=\cfrac{\cfrac{12}{5}-0}{3+4}(x+4)$
or, $35y=12x+48$
or, $-12x+35y=48$
and another equation of reflected ray$\Rightarrow y=\cfrac{\cfrac{-12}{5}-0}{3+4}(x+4)$
or, $35y=-12x-48$
or, $-12x-35y=48$

The tangent and normal to the ellipse $x^2\, +\, 4y^2\, =\, 4$ at a point $P(\theta)$ on it meet the major axis in $Q$ and $R$ respectively. If $QR = 2$, the eccentric angle $\theta$ of $P$ is given by 

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

  1. $\cos \theta\, =\, \pm\, \dfrac23$

  2. $\sin \theta\, =\, \pm\, \dfrac23$

  3. $\tan \theta\, =\, \pm\, \dfrac23$

  4. $\cot \theta\, =\, \pm\, \dfrac23$

Show answer
Correct Option: A
Explanation:

The equation of tangent to ellipse is $bx cos\theta+ay sin\theta=ab$

It meets major axis at $x=\dfrac{a}{cos\theta}$
The equation of normal to ellipse is $ax sec \theta -by cosec \theta=a^{2}-b^{2}$
It meets the major axis at $x=\dfrac{a^{2}-b^{2}}{a sec\theta}$
Here $a=2$ and $b=1$
So, point $Q$ is $\left(\dfrac{2}{cos\theta},0 \right)$ and point $R$ is $\left(\dfrac{3}{2 sec\theta},0\right)$
The distance between them $QR= \dfrac{3cos\theta}{2}-\dfrac{2}{cos\theta}=2$
$\Rightarrow 3\cos ^{ 2 }{ \theta  } -4cos \theta-4=0$
$\Rightarrow cos \theta=-\dfrac{2}{3}$

The interior angle of a regular polygon is double the exterior angle. Then the number in the polygon is 

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $6$

  2. $8$

  3. $9$

  4. None of these

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Correct Option: A
Explanation:

Let the number of sides in the polygon is $n$

Let the measure of exterior angles be $x$ respectively.
$\therefore$   Measure of interior angle $=2x$
So,
$\Rightarrow$  $n\times 2x=(2n-4)\times 90^o$
$\Rightarrow$  $nx=(n-2)\times 90^o$               ----- ( 1 )
Again we know that,
$\Rightarrow$  $nx=360^o$
$\Rightarrow$  $(n-2)\times 90^o=360^o$               [ From ( 1 ) ]
$\Rightarrow$  $n-2=4$
$\Rightarrow$  $n=6$
$\therefore$   The number of sides in the polygon are $6.$

State true or false:
Is it possible to have a regular polygon whose each exterior angle is $\displaystyle 32^{\circ}$
maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: B
Explanation:
Each exterior angle of regular polygon of n sides is given by $\dfrac{360}{n}$ degree
$\text{According to the question}$
$\dfrac{360}{n} degree=32^0$
$\Rightarrow n=\dfrac{360}{32}$
$\Rightarrow n=\dfrac{45}4$
$\text{Which is not possible because number of sides can never be in fraction.}$

State true or false:
Is it possible to have a regular polygon whose each exterior angle is $\displaystyle 20^{\circ}$

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
$\text{Each exterior angle of regular polygon of n sides is given by}\dfrac{360}{n} degree$
$\text{According to question}$
$\dfrac{360}{n} degree=20^0$
$\Rightarrow n=\dfrac{360}{20}$
$\Rightarrow n=18$
$\text{Clearly there is a polygon of sides 18 whose each exterior angle is 20}^0$

Two alternate sides of a regular polygon, when produced, meet at a right angle, then find the value of each exterior angle of the polygon.

maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

  1. $\displaystyle 45^{\circ}$

  2. $\displaystyle 32^{\circ}$

  3. $\displaystyle 62^{\circ}$

  4. $\displaystyle 15^{\circ}$

Show answer
Correct Option: A
Explanation:

$We\quad know\quad External\quad angles\quad of\quad a\quad regular\quad polygon\quad are\quad equal.$


$\ When\quad the\quad two\quad alternate\quad sides\quad of\quad a\quad regular\quad polygon\quad are\quad produced,\quad they\quad meet\quad at\quad right\quad angle.$

$\ These\quad two\quad extended\quad sides\quad form\quad a\quad triangle\quad with\quad the\quad side\quad of\quad the\quad polygon\quad in\quad between.$

$\ Sum\quad of\quad all\quad interior\quad angles\quad of\quad a\quad \triangle ={ 180 }^{ o }$

$\ \Rightarrow 2\times External\quad angle\quad +\quad { 90 }^{ o }\quad =180$

$\ \Rightarrow 2\times External\quad angle=180-90=90$

$\ \Rightarrow External\quad angle=\dfrac { 90 }{ 2 }$

$ \ \Rightarrow External\quad angle={ 45 }^{ o }
$