Tag: maths

Questions Related to maths

Point $\left(\sqrt5, \dfrac4{\sqrt5}\right)$ lies _____ the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{4} = 1$.

position of point wrt ellipse ellipse maths

  1. inside

  2. outside

  3. on

  4. None of the above

Show answer
Correct Option: C
Explanation:

The point $(x,y)$ lies inside the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if it satisfies the following condition i.e.

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leq 1$
To check $\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$ lies in the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{4} = 1$
Here $a=2$ and $b=5$
Take $(x,y)=\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$
$\therefore$ $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{(\sqrt5)^2}{4}+\dfrac{(\frac{4}{\sqrt5})^2}{25}=\dfrac{5}{4}+\dfrac{16}{5\times 25}=\dfrac{689}{600} < 1$
Hence, $\left(\sqrt{5}, \dfrac{4}{\sqrt{5}}\right)$ lies in the given ellipse.

State the following statement is true or false
$(3,0),(0,4)$ and $(4,0)$ are all points that lie on the ellipse.

position of point wrt ellipse ellipse maths

  1. True

  2. False

Show answer
Correct Option: B

Determine position of a point $(2,3)$ with respect to the ellipse $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{25}=1$.

position of point wrt ellipse ellipse maths

  1. Outside

  2. Inside

  3. On the ellipse

  4. None of the above

Show answer
Correct Option: B
Explanation:

Given equation is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1$

$ \dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1=0$

The curve is defined by, 

$f(x,y)=\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { { (2) }^{ 2 } }{ 16 } +\dfrac { { (3) }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { 1 }{ 4 } +\dfrac { 9 }{ 25 } -1=\dfrac { 25+36-100 }{ 100 } =-\dfrac { 39 }{ 100 } \\ f(2,3)<0$

So, the point lies inside the ellipse.

Option B is correct.

Position of a point $(3,-4))$ with respect to the ellipse $16x^{2}+9y^{2}=144$ lies 

position of point wrt ellipse ellipse maths

  1. outside

  2. inside

  3. on

  4. None of the above

Show answer
Correct Option: A
Explanation:

Given equation is $16{ x }^{ 2 }+9{ y }^{ 2 }=144$

$ f(x,y)=16{ x }^{ 2 }+9{ y }^{ 2 }-144\ f(3,-4)=16{ (3) }^{ 2 }+9(-4)^{ 2 }-144\ f(3,-4)=144+144-144\ f(3,-4)=144\ f(3,-4)>0$
So, the point lies outside the ellipse.

Option A is correct.

The position of point $(4,3)$ with respect to the ellipse $\dfrac{x^2}{4}+\dfrac {y^2}{3}=1$

position of point wrt ellipse ellipse maths

  1. Inside 

  2. Outside

  3. On the Ellipse 

  4. None.

Show answer
Correct Option: B
Explanation:

Given ellipse $\dfrac {x^2}{4}+\dfrac {y^2}{3}=1$


Given point $(4,3)$

The position of point is $\dfrac {4^2}{4}+\dfrac {3^2}3-1\4+3-1=6>0$
So the point s outside the ellipse 

The distance of a point P on the ellipse $\dfrac{{{x^2}}}{{12}} + \dfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the ecentric angle of P is

position of point wrt ellipse ellipse maths

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

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

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

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

Show answer
Correct Option: A,D
Explanation:

Ellipse$:\cfrac { x^{ 2 } }{ 12 } +\cfrac { y^{ 2 } }{ 4 } =1$

Centre$:C(0,0)$
Given that $PC=\sqrt { 6 } $
Parametric coordinates $:y=b\sin { \phi  } ,x=a\cos { \phi  } $
$\phi$ is eccentric angle. 
$\Rightarrow 6=(2\sqrt { 3 } \cos { \phi  } -0)^{ 2 }+(2\sin { \phi  } -0)^{ 2 }$
$ \Rightarrow 6=12\cos { ^{ 2 }\phi  } +4\sin { ^{ 2 }\phi  } $
$ \Rightarrow 2=8\cos ^{ 2 }{ \phi  } $
$ \Rightarrow \cos ^{ 2 }{ \phi  } =\cfrac { 1 }{ 4 } $
$ \Rightarrow \cos { \phi  } = \pm \cfrac { 1 }{ 2 } $
$ \Rightarrow \phi = \cfrac { \pi  }{ 3 } $
$\Rightarrow \phi =\cfrac { 2\pi  }{ 3 } $

Evaluate $\displaystyle \int x^2+3x+5\ dx$ 

position of point wrt ellipse ellipse maths

  1. $\dfrac {x^3}3+3\dfrac {x^2}3+\dfrac 53+c$

  2. $\dfrac {x^2}2+ 3x+5+c $

  3. $ \dfrac {x^3}3+3\dfrac {x^2}2+5x+c $

  4. None of the above.

Show answer
Correct Option: C
Explanation:

$\displaystyle \int x^2+3x+5\ dx$ 


$=\displaystyle \int x^2 dx+\int 3x dx+\int 5 dx$


$=\dfrac {x^3}{3}+\dfrac {3x^2}{2}+5x+C$


An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $\dfrac 23$ then the eccentricity of the ellipse is  

position of point wrt ellipse ellipse maths

  1. $\dfrac{2\sqrt{2}}{3}$

  2. $\sqrt{5}$

  3. $8$

  4. $2$

Show answer
Correct Option: A
Explanation:
Probability  $p=\dfrac 23$

Let the radius of circle$=a$

Major axis$=2a$

Minor axis$=2b$

Area of circles$=\pi a^2$

Area of ellipse$=\pi ab$

as because

$P=\dfrac{\pi ab-\pi a^2}{\pi ab}=1-\dfrac{a}{b}$

$\dfrac{a}{b}=1-\dfrac{2}{3}=\dfrac{1}{3}$

$e=\sqrt{1-\left(\dfrac{a}{b}\right)^2}-\sqrt{1-\left(\dfrac{1}{3}\right)^2}$

$=\sqrt{\dfrac{8}{9}}$

Eccentricity$=e=\dfrac{2\sqrt{2}}{3}$.

The position of the point (1,3)n with respect to the ellipse $4x^{2}+9y^{2}-16x-54y+61=0$ is 

position of point wrt ellipse ellipse maths

  1. Outside the ellipse 

  2. On the ellipse

  3. On the major axis

  4. On the minor axis

Show answer
Correct Option: A
Explanation:

We have,

Equation of given ellipse,

$4{{x}^{2}}+9{{y}^{2}}-16x-54y+61=0$

Given point,

Let, $\left( x,y \right)=\left( 1,\,3 \right)$

Then, position of point in given ellipse is

$ 4{{\left( 1 \right)}^{2}}+9{{\left( 3 \right)}^{2}}-16\left( 1 \right)-54\left( 3 \right)+61=0 $

$ \Rightarrow 4+81-16-162+61=0 $

$ \Rightarrow 146-178=0 $

$ \Rightarrow -32<0 $

The position of point outside the ellipse.

If the point $(a\sin\theta, a\cos\theta)$ lies on the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ then the value of $\sin 2\theta$ is (where $a\neq b, a>0, b>0$ and $e$ is the eccentricity of the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$)

position of point wrt ellipse ellipse maths

  1. $\dfrac{2\sqrt{1+e^{2}}}{2+e^{2}}$

  2. $\dfrac{2\sqrt{1-e^{2}}}{2+e^{2}}$

  3. $\dfrac{2\sqrt{1-e^{2}}}{2-e^{2}}$

  4. $\dfrac{2\sqrt{1+e^{2}}} {2-e^{2}}$

Show answer
Correct Option: A