Tag: mathematics and statistics

Questions Related to mathematics and statistics

The equation of the ellipse having vertices at $\displaystyle \left( \pm 5,0 \right) $ and foci $\displaystyle \left( \pm 4,0 \right) $ is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  2. $\displaystyle 9{ x }^{ 2 }+25{ y }^{ 2 }=225$

  3. $\displaystyle \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 25 } =1$

  4. $\displaystyle 4{ x }^{ 2 }+5{ y }^{ 2 }=20$

Show answer
Correct Option: B
Explanation:

The vertices and foci of an ellipse are $\displaystyle \left( \pm 5,0 \right) $ and $\displaystyle \left( \pm 4,0 \right) $ respectively.
$\displaystyle \therefore \quad a=5$ and $\displaystyle ae=4$
$\displaystyle \Rightarrow \quad e=\frac { 4 }{ 5 } $
We know that,
$\displaystyle e=\sqrt { 1-\frac { { b }^{ 2 } }{ { a }^{ 2 } }  } $
$\displaystyle \Rightarrow \quad \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 25 } \Rightarrow { b }^{ 2 }=9$
Hence, equation of an ellipse is
$\displaystyle \frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow 9{ x }^{ 2 }+25{ y }^{ 2 }=225$

The sum of the focal distances of any point on the conic $\dfrac {x^{2}}{25} + \dfrac {y^{2}}{16} = 1$ is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $10$

  2. $9$

  3. $41$

  4. $18$

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Correct Option: A
Explanation:

We know, if P is any point on the curve, then Sum of focal distances $=$ length of major axis
i.e., $SP + S'P = 2a$
$= 2(5) [\because a^{2} = 5^{2}]$
$= 10$

The graph of the equation $x^2+\dfrac{y^2}{4}=1$ is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. an ellipse

  2. a circle

  3. a hyperbola

  4. a parabola

  5. two straight lines

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Correct Option: A
Explanation:

Given, ${x}^{2}+\dfrac{{y}^{2}}{4}=1$
It is in the form of  $ \dfrac{{x}^{2}}{{a}^{2}}+\dfrac{{y}^{2}}{{b}^{2}}=1$
Therefore, it represents an ellipse.

The graph of the equation $4y^2 + x^2= 25$ is

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. a circle

  2. an ellipse

  3. a hyperbola

  4. a parabola

  5. a straight line

Show answer
Correct Option: B
Explanation:

Given, $4{y}^{2}+{x}^{2}=25$

$\Rightarrow \dfrac { { y }^{ 2 } }{ 25/4 } +\dfrac { { x }^{ 2 } }{ 25 } =1$
It is in the form of ellipse $\left (\dfrac { { y }^{ 2 } }{ {a}^{2} } +\dfrac { { x }^{ 2 } }{ {b}^{2} } =1\right)$
So, the correct answer is option $B$.

Latus rectum of the conic satisfying the differential equation $x dy+y dx=0$ and passing through the point $(2,8)$ is :

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $4\sqrt{2}$

  2. $8$

  3. $8\sqrt{2}$

  4. $16$

Show answer
Correct Option: C
Explanation:

The differential equation is

$xdy+ydx=0$
$d(xy)=0$
By integrating we get,
$xy=k$..........(1)

Equation (1) passes through the point $(2,8)$, so
$(2)(8)=k$
$k=16$

So, equation of the conic is 
$xy=16$
which is a rectangular hyperbola $(xy=c^{2})$, where $c=4$

Length of latus rectum for rectangular hyperbola is $2\sqrt{2}c=8\sqrt{2}$
 





The foci of an ellipse are located at the points $(2, 4)$ and $(2, -2)$. The points $(4, 2)$ lies on the ellipse. If $a$ and $b$ represent the lengths of the semi-major and semi-minor axes respectively, then the value of $(ab)^{2}$ is equal to

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $68 + 22\sqrt {10}$

  2. $6 + 22\sqrt {10}$

  3. $26 + 10\sqrt {10}$

  4. $6 + 10\sqrt {10}$

Show answer
Correct Option: C
Explanation:

The distance between the foci is $6$, so $c = 3$.
The sum of the distance from $(4, 2)$ to each of the foci is the major axis length,
so
$2a = \sqrt {(4 - 2)^{2} + (2 - 4)^{2}} + \sqrt {(4 - 2)^{2} + (2 + 2)^{2}}$
$= \sqrt {4 + 4} + \sqrt {4 + 16} = \sqrt {8} + \sqrt {20}$
$= 2\sqrt {2} + 2\sqrt {5} \Rightarrow a = \sqrt {2} + \sqrt {5}$
Also, for an ellipse,
$b^{2} = a^{2} - c^{2} = (\sqrt {2} + \sqrt {5})^{2} - 3^{2}$
$= 7 + 2\sqrt {10} = -2 + 2\sqrt {10}$.
Thus, we have
$(ab)^{2} = (7 + 2\sqrt {10})(-2 + 2\sqrt {10})$
$= -14 + 14\sqrt {10} - 4\sqrt {10} + 40$
$= 26 + 10\sqrt {10}$.

Which of the following is/are not false?

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. The mid point of the line segment joining the foci is called the centre of the ellipse.

  2. The line segment through the foci of the ellipse is called the major axis.

  3. The end points of the major axis are called the vertices of the ellipse.

  4. Ellipse is symmetric with respect to Y-axis only.

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

(A) Midpoint of the line segment joining the foci is called the centre of ellipse: TRUE


(B) Line segment through the foci is called major axis: TRUE

(C) End point of major axis are called vertices of ellipse: TRUE

(D) Ellipse is symmetric with respect to Y-axis only.: FALSE
(Ellipse is symmetric to both x-axis and y-axis)

The equation $2x^2+3y^2-8x-18y+35=\lambda$ represents?

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. A circle for all $\lambda$

  2. An ellipse if $\lambda < 0$

  3. The empty set if $\lambda > 0$

  4. A-point if $\lambda = 0$

Show answer
Correct Option: D
Explanation:

Given:

$ 2x^{2} + 3y - 8x - 18y + 35 = \lambda  $

$ 2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda  $

$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = \lambda  $

For $ \lambda = 0 $, then

$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = 0  $

Thus, the point is $ \left ( 2,3 \right ) $.

Hence, the correct option is ‘d’.

The equation of ellipse whose major axis is along the direction of x-axis, eccentricity is $e=2/3$

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

  1. $36x^2+20y^2=405$

  2. $20x^2+36y^2=405$

  3. $30x^2+22y^2=411$

  4. $22x^2+32y^2=409$

Show answer
Correct Option: B
Explanation:
$\begin{array}{l} e=\frac { 2 }{ 3 } =\sqrt { \frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ c }  }  \\ \Rightarrow { \left( { \frac { 2 }{ 3 }  } \right) ^{ 2 } }=\frac { { { a^{ 2 } }-{ b^{ 2 } } } }{ { { a^{ 2 } } } }  \\ \Rightarrow \frac { { 4{ a^{ 2 } } } }{ a } ={ a^{ 2 } }-{ b^{ 2 } } \\ \Rightarrow { b^{ 2 } }={ a^{ 2 } }-4{ a^{ 2 } }=\frac { { 5{ a^{ 2 } } } }{ 9 } \to \left( i \right)  \end{array}$
Equation of Ellipse are
$\begin{array}{l} \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { { y^{ 2 } } } }{ { { b^{ 2 } } } } =1 \\ \Rightarrow \frac { { { x^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { 9{ y^{ 2 } } } }{ { 5{ a^{ 2 } } } } =1\to \left( { ii } \right)  \\ Put\, \, { a^{ 2 } }=\frac { { 405 } }{ { 20 } } \, \, \left( { From\, \, option\, \, in\, \, equation\left( i \right)  } \right)  \\ Then,\, \, { b^{ 2 } }=\frac { { 405 } }{ { 360 } }  \end{array}$
Hence, equation of ellipse is
$ \Rightarrow \frac{{{x^2}}}{{\left( {\frac{{405}}{{20}}} \right)}} + \frac{{{y^2}}}{{\left( {\frac{{405}}{{36}}} \right)}} =  - 1,20{x^2} + 36{y^2} = 405$

Eccentricity of ellipse $\frac{{{x^2}}}{{{a^2} + 1}} + \frac{{{y^2}}}{{{a^2} + 2}} = 1\,is\,\frac{1}{{\sqrt 3 }}$ then length of Latus rectum is 

mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

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

  2. $\frac{4}{{\sqrt 3 }}$

  3. $2\sqrt 3 $

  4. $\frac{{\sqrt 3 }}{2}$

Show answer
Correct Option: B
Explanation:
Let ${ A }^{ 2 }={ a }^{ 2 }+1$,  ${ B }^{ 2 }={ a }^{ 2 }+2$    {Here, ${ B }^{ 2 }>{ A }^{ 2 }$}
So, $e=\sqrt { 1-\dfrac { { A }^{ 2 } }{ { B }^{ 2 } }  } =\sqrt { 1-\dfrac { { a }^{ 2 }+1 }{ { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 1 }{ \sqrt { 3 }  } $
So,  $\sqrt { { a }^{ 2 }+2 } =\sqrt { 3 } \Rightarrow a=\pm 1$
So, length of lotus return $=2\dfrac { { A }^{ 2 } }{ B } $
Length $=\dfrac { 2\left( { a }^{ 2 }+1 \right)  }{ \sqrt { { a }^{ 2 }+2 }  } =\dfrac { 2\left( 2 \right)  }{ \sqrt { 3 }  } =\dfrac { 4 }{ \sqrt { 3 }  } $