Tag: maths

Questions Related to maths

Let $E$ be the ellipse $\displaystyle \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 4 } =1$ and $C$ be the circle ${ x }^{ 2 }+{ y }^{ 2 }=9$. Let $P$ and $Q$ be the points $(1,2)$ and $(2,1)$ respectively. Then

position of point wrt ellipse ellipse maths

  1. $Q$ lies inside $C$ but outside $E$

  2. $Q$ lies outside both $C$ and $E$

  3. $P$ lies inside both $C$ and $E$

  4. $P$ lies inside $C$ but outside $E$

Show answer
Correct Option: D
Explanation:

Since $\displaystyle \frac { { 1 }^{ 2 } }{ 9 } +\frac { { 2 }^{ 2 } }{ 4 } -=\frac{1}{9}>0$

$\therefore P(1,2)$ lies outside $E$
Since $\displaystyle \frac { { 2 }^{ 2 } }{ 9 } +\frac { { 1 }^{ 2 } }{ 4 } -1<0$
$\therefore Q(2,1)$ lies inside $E$
Since ${ 1 }^{ 2 }+{ 2 }^{ 2 }-9<0$
$\therefore P(1,2)$ lies inside $C$
Since ${ 2 }^{ 2 }+{ 1 }^{ 2 }-9<0$
$\therefore Q(2,1)$ also lies inside $C$
$\therefore P$ lies inside $C$ but outside $E$.

Find the equation of the ellipse whose eccentricity is $\dfrac{4}{5}$ and axes are along the coordinate axes and foci at $(0, \pm 4)$.

position of point wrt ellipse ellipse maths

  1. $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$

  2. $\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

  3. $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$

  4. $\dfrac{x^2}{9}+\dfrac{y^2}{36}=1$

Show answer
Correct Option: A
Explanation:

Let the required equation of the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.


According to the problem, the coordinates of the foci are $(0, \pm 4)$.

We know, coordinates of foci are $(0, \pm be)$.

Therefore, $be =4$

$b\left (\dfrac{4}{5}\right )=4$

$b=5$

$b^2=25$

Now, $a^2=b^2(1-e^2)$

$a^2=5^2\left (1-\dfrac{16}{25}\right )$

$a^2=9$

Thus, the required equation of ellipse is $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$.

The point $(4, -3)$ with respect to the ellipse $4x^2+5y^2=1$.

position of point wrt ellipse ellipse maths

  1. lies on the curve

  2. lies inside the curve

  3. lies outside the curve

  4. lies focus of the curve

Show answer
Correct Option: C
Explanation:

$Equation\quad of\quad ellipse:\quad 4{ x }^{ 2 }+5{ y }^{ 2 }=1\ Putting\quad the\quad point\quad (4,-3)\quad on\quad the\quad ellipse\quad we\quad get:\ \quad =4{ (4) }^{ 2 }+5{ (-3) }^{ 2 }-1\ =\quad 64+45-1=28>0\ \therefore \quad The\quad point\quad lies\quad outside\quad the\quad ellipse.$


Option [C]

Consider the ellipse with the equation $x^{2}+3y^{2}-2x-6y-2=0.$ The eccentric angle of a point on the ellipse at a distance 2 units from the contra of the ellipse is

position of point wrt ellipse ellipse maths

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

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

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

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

Show answer
Correct Option: A
Explanation:

We have,  $x^{2}+3y^{2}-2x-6y-2=0.$

$\Rightarrow$ $x^{2}-2x+1-1+3y^{2}-6y+3-3-2=0.$

$\Rightarrow$ $(x-1)^{2}+3(y-1)^{2}=6$

$\Rightarrow$ $(x-1)^{2}+3(y-1)^{2}=6$ which becomes $x^{2}+3y^{2}=6$ on shifting the origin to $(1, 1)$. Any point with eccentric angle $\theta $ is $(\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$ 

$\Rightarrow$ $4=6cos^{2}\theta +2sin^{2}\theta \Rightarrow 4cos^{2}\theta =2\Rightarrow cos\theta =\pm \dfrac{1}{\sqrt{2}}$

$\Rightarrow$ Hence $\theta =\dfrac{\pi }{4}$ 

Find the set of value(s) of $\alpha$ for which the point $\left ( 7\,-\, \displaystyle \frac{5}{4}\alpha,\,\alpha \right )$ lies inside the ellipse $\displaystyle \frac{x^2}{25}\,+\,\frac{y^2}{16}\,=\, 1.$

position of point wrt ellipse ellipse maths

  1. $ \displaystyle\left( \frac{17}{5} \dfrac{12}{5}\right) $

  2. $ \left(\dfrac{12}{5},\dfrac{16}{5}\right) $

  3. $ \dfrac{-16}{5} $

  4. None of these

Show answer
Correct Option: B
Explanation:

$\displaystyle \frac{x^2}{25}\,+\,\frac{y^2}{16}\,=\, 1$


Since point $(7\,-\, \displaystyle \frac{5}{4}\alpha,\,\alpha)$ lies inside the ellipse

$\therefore S _1\,<\,0$

$\Rightarrow 16 (7\,-\, \displaystyle \frac{5}{4}\alpha)^2\,+\, 25.\alpha^2\,<\,400$

$\Rightarrow\,(28\, -\,5\alpha)^2\,+\, 25\alpha^2\,<\, 400$

$\Rightarrow\, 50\alpha^2\, -\, 280\,\alpha\,+\, 384\, < \,0$

$\Rightarrow\, 25\alpha^2\, -\, 140\,\alpha\,+\, 192\, < \,0$

$\Rightarrow (5\alpha-12)(5\alpha-16)<0$

$\Rightarrow \, \alpha \, \in \, \left( \dfrac { 12 }{ 5 } ,\, \dfrac { 16 }{ 5 }  \right) $

$\mathrm{A}$ssertion ($\mathrm{A}$): The point $(5,-2)$ lies outside the ellipse $24x^{2}+7y^{2}=12$.
Reason (R): lf the point $(x _{1},y _{1})$ lie outside the ellipse $\mathrm{S}=0$ then $S _{11}>0$ 

position of point wrt ellipse ellipse maths

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

  2. Both A and R are true but R is not coorect explanation of A

  3. A is true but R is false

  4. A is false but R is true

Show answer
Correct Option: A
Explanation:

$24x^{2}+7y^{2}-12=5$

$S(5,-2)=24\times 25+7\times 4-12$

$S(5,-2)>0$

$\therefore $ lies outside the ellipse.
$S=0$
$S(x _{1}y _{1})>0$ the point is outside the ellipse.

The point $(2\cos \theta , 3\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$.

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on the periphery of

  4. on the auxillary of

Show answer
Correct Option: C
Explanation:

Given point is $\left ( 2\cos\theta, 3\sin\theta  \right )$


Substituting given point in the ellipse equation:
$\dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}-1$
$= \dfrac{4\cos^{2}\theta}{4}+\dfrac{9\sin^{2}\theta }{9}-1$
$= \cos^{2}\theta + \sin^{2}\theta -1$
$= 1-1=0$

$\therefore$ Given point satisfies the ellipse 
$\Rightarrow$ point lie on periphery of ellipse.

The distance of point '$\theta$' on the ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1$ from a focus is:

position of point wrt ellipse ellipse maths

  1. $a(e + \cos \theta)$

  2. $a(e - \cos \theta)$

  3. $a(1 + e \cos \theta)$

  4. $a(1 + 2e \cos \theta)$

Show answer
Correct Option: C
Explanation:
Given equation of ellipse is
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
Any point on the ellipse will be $(a cos\theta, b sin \theta)$
$P=(acos\theta, b sin \theta)$
Centre of the ellipse is $(0,0)$
Ellipse is parallel to horizontal axis
Foci of the ellipse is
$F=(h-ae,k)$ if $(h,k) $ is the centre
$F=(0-ae,0)=(-ae,o)$
Distance $FP=\sqrt{(-ae-cos\theta)^2+(bsin\theta-0)^2}$
$=a \sqrt{(e^2+cos^2 \theta+2ecos\theta+(1-e^2)sin \theta-0)^2}$
$a\sqrt{(e^2+cos^2 \theta+2ecos\theta+sin^2 \theta-e^2sin^2\theta)}$
$a\sqrt{(1+2ecos\theta+e^2(1-sin^2\theta)}$
$a\sqrt{(1+2ecos \theta+e^2cos^2\theta)}$
$a\sqrt{(1+ecos\theta)^2}$
$FP=a(1+ecos\theta)$



$(2,3)$ lies _______  the ellipse $16 x^{2} + 9y^{2} - 16x - 32 = 0$

position of point wrt ellipse ellipse maths

  1. Inside

  2. Outside

  3. On

  4. None of the above

Show answer
Correct Option: B
Explanation:
Given ellipse $16x^2+9y^2-16x-32=0$
Let $S=16x^2+9y^2-16x-32=0$
Put $(2,3) $ in $S$ we get
$S(2,3)=16(2)^2+9(3)^2-16(2)-32=81 $
$S(2,3)>0$
So it $(2,3) $ lies outside the ellipse.

The point $(4\cos \theta , 4\sin \theta)$ lies ____________ the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on the periphery

  4. on the auxiliary circle

Show answer
Correct Option: D
Explanation:

Equation of ellipse is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 9 } =1$

Equation of auxiliary circle of ellipse is
${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\ { x }^{ 2 }+{ y }^{ 2 }=16$
Substiuting $(4\cos\theta,4\sin\theta$) in the equation, we get
${ (4\cos { \theta  } ) }^{ 2 }+{ (4\sin { \theta  } ) }^{ 2 }=16\ 16(\cos ^{ 2 }{ \theta  } +\sin ^{ 2 }{ \theta  } )=16\ 16=16$
The point satisfies the equation of auxiliary circle.
So, option D is the correct.