Tag: maths

Questions Related to maths

If $P=(x, y), F _1=(3, 0), F _2=(-3, 0)$ and $  16x^2+25y^2=400$, then $PF _1+PF _2$ equals

position of point wrt ellipse ellipse maths

  1. $8$

  2. $6$

  3. $10$

  4. $12$

Show answer
Correct Option: C
Explanation:

Given equation is $16x^2+25y^2=400$
The ellipse can be written as, $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$
Here $a^2=25, b^2=16$,  but  $b^2=a^2(1-e^2)$
$\Rightarrow \dfrac {16}{25}=1-e^2$
$\Rightarrow e^2=1-\dfrac {16}{25}=\dfrac {9}{25} $
$ \Rightarrow e=\pm \dfrac {3}{5}$
Foci of the ellipse are $(+ae, 0)$ $=$ $ (\pm 3, 0)$, i.e., $F _1$ and $F _2$.
We have $PF _1+PF _2=2a=10$ for every point $P$ on the ellipse.

The position of the point $(1, 2)$ relative to the ellipse $2x^{2} + 7y^{2} = 20$ is

position of point wrt ellipse ellipse maths

  1. outside the ellipse

  2. inside the ellipse but not at the focus

  3. on the ellipse

  4. at the focus

Show answer
Correct Option: A
Explanation:

$f\left( x,y \right) ={ 2x }^{ 2 }+7{ y }^{ 2 }-20$

Point will be outside of ellipse if $f\left( 1,2 \right) >0$

on ellipse if $f\left( 1,2 \right) =0$

will be inside if $f\left( 1,2 \right) <0$

$ f\left( 1,2 \right) ={ 2\times 1 }^{ 2 }+7{ \times 2 }^{ 2 }-20=10$

Here we see that $f\left( 1,2 \right) >0$ so point will be outside of ellipse

So correct answer will be option A

The minimum distance of origin from the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is $(a>0,b>0)$

position of point wrt ellipse ellipse maths

  1. a-b

  2. a+b

  3. 2a+2b

  4. 2(a-b)

Show answer
Correct Option: B
Explanation:
Let $(a sec\theta,b cosec \theta)$ be a point on the curve, then its diastance  from the origin
$=\sqrt{a^2 sec^2 \theta + b^2 cosec ^2 \theta}$
$\therefore f(\theta)=a^2 sec^2 \theta +b^2 cosec ^2 \theta$
$=a^2+a^2 tan ^2 \theta + b^2 + b^2 cot ^ 2 \theta$
$=a^2+b^2 + a^2 tan ^2 \theta + b^2 cot ^2 \theta$
$\geq a^2+b^2+2 \sqrt{a^2 b^2}=(a+b)^2$
$\therefore$ minimum value of $f(\theta)=(a+b)^2$
$\therefore$ minimum value of $\sqrt{a^2 sec^2 \theta + b^2 cosec^2 \theta }$ is $a+b$

If $a$ and $c$ positive real number and the ellipse $\dfrac { { x }^{ 2 } }{ { 4c }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { c }^{ 2 } } =1$ has four distinet points in common with the circle ${ x }^{ 2 }+{ y }^{ 2 }=9{ a }^{ 2 }$, then

position of point wrt ellipse ellipse maths

  1. $6ac+9{ a }^{ 2 }-2{ c }^{ 2 }>0$

  2. $6ac+9{ a }^{ 2 }-2{ c }^{ 2 }<0$

  3. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }<0$

  4. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }>0$

Show answer
Correct Option: 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 $2/3$ then the eccentricity of the ellipse is:

position of point wrt ellipse ellipse maths

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

  2. $\dfrac{\sqrt{5}}{3}$

  3. $\dfrac{8}{9}$

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

Show answer
Correct Option: A

The segment of the tangent at the point P to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, intercepted by the auxiliary circle subtends a right angle at the origin. If the eccentricity of the ellipse is smallest possible, then the point P can be

position of point wrt ellipse ellipse maths

  1. $(0,ae)$

  2. $(a,0)$

  3. $(-a,0)$

  4. $(0,-b)$

Show answer
Correct Option: D
Explanation:

Equation of tangent at $P (a cos \theta, b sin \theta)$ is
$\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$
$\therefore$ Equation of the lines OA and OB is
$x^{2}+y^{2}-a^{2}\left(\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}\right)^{2}=0$
since the lines OA and OB are perpendicular
$\therefore$ $1- \cos ^{2}\theta + 1 - \frac{a^{2}}{b^{2}} \sin^{2}\theta=0$
i.e $\sin^{2}\theta + 1-\frac{a^{2}}{b^{2}} \sin^{2}\theta$
i.e $\frac{b^{2}}{a^{2}}=\frac{sin^{2}\theta}{1+ \sin^{2} \theta}$
$\therefore e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{\sin^{2} \theta}{1+\sin^{2}\theta}=\frac{1}{1+\sin^{2}\theta}$
$\therefore$ e is least if $\theta = \pm \frac{\pi}{2}$
$\therefore$ the point P is $(0,\pm b)$

If one end of the diameter of the ellipse $4x^2+y^2=16$ is $(\sqrt 3, 2)$, then the other end is:

position of point wrt ellipse ellipse maths

  1. $(\sqrt 3, 2)$

  2. $(-\sqrt 3, 2)$

  3. $(-\sqrt 3, -2)$

  4. $(\sqrt 3, -2)$

Show answer
Correct Option: C
Explanation:

The center of ellipse is $(0,0)$

The diameter of ellipse will pass through the center of ellipse and center will divide the diameter into two halfs
One end of diameter is $(\sqrt3,2) $ , Let the other end be $(h,k)$
We have $h+\sqrt3=0$ and $k+2=0$
$\Rightarrow h=-\sqrt3,k=-2$
Therefore option $C$ is correct

The point P on the ellipse $4x^2+9y^2=36$ is such that the area of the $\Delta PF _1F _2=\sqrt{10} Sq$ units, where $F _1.F _2$ are Foci. Then P has the coordinates

position of point wrt ellipse ellipse maths

  1. $(\pm\dfrac{3}{\sqrt{2}},\sqrt{2})$

  2. $(\dfrac{3}{2},2)$

  3. $(\dfrac{-3}{2},-2)$

  4. NONE

Show answer
Correct Option: A
Explanation:

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

So, $a=3,b=2$

In an ellipse, Focal points, $F=(\pm\sqrt {a^2-b^2},0)=(\pm\sqrt 5,0)$
So, $F _1F _2=2\sqrt 5$ which is the base of the triangle $PF _1F _2$

An arbitrry point P on ellipse is $(acos\theta,b\sin\theta)$
So, $Height\ of\ \triangle PF _1F _2=b\sin\theta=2\sin\theta$

ie, $Area=\dfrac 12base\times height=\dfrac 12\times 2\sqrt 5\times2\sin\theta=2\sqrt 5\sin\theta=\sqrt {10}$
$\Rightarrow \sin\theta=\dfrac 1{\sqrt 2}$
$\Rightarrow \cos\theta=\pm\dfrac 1{\sqrt 2}$
And $P=(a\cos\theta,b\sin\theta)=(\pm\dfrac 3{\sqrt 2},\sqrt 2)$

So, Option$ A$ has one of the anwers.

Which of the following is an (x,y) coordinate pair located on the ellipse $4x^2 + 9y^2 = 100$?

position of point wrt ellipse ellipse maths

  1. $(1, 3.5)$

  2. $(1.4, 3.2)$

  3. $(1.9, 2.9)$

  4. $(2.3, 3.1)$

  5. $(2.7, 2.6)$

Show answer
Correct Option: B
Explanation:
  • Substitute each and every point on given ellipse and see which will satisfy the equation
  • Take point $(1.4,3.2)$ , when we substitute we get $4(1.96)+9(10.24) = 100$
  • Therefore correct answer is option $B$

If $iz^4 + 1 = 0$, then z can take the value

de moivre’s theorem and its applications demoivre's theorem complex numbers maths

  1. $\displaystyle \frac{1 + i}{\sqrt 2}$

  2. $\cos \displaystyle \frac{\pi}{8} + i \sin \frac{\pi}{8}$

  3. $\displaystyle \frac{1}{4 i}$

  4. $i$

Show answer
Correct Option: B
Explanation:
$iz^4 + 1 = 0 \Rightarrow z^4 = - \displaystyle \frac{1}{i} = \frac{i^2}{i} = i$
Let $z^4 = \cos  \displaystyle \frac{\pi}{2} + i  \sin  \frac{\pi}{2}$
$\therefore z = \displaystyle \left [ \cos  \frac{\pi}{2} + i  \sin  \frac{\pi}{2} \right ]^{1/4}$
U\sin g De-Moivre's theorem,
$(\cos   \theta + i  \sin  \theta)^n = \cos   n \theta + i  \sin   n \theta, n  \varepsilon I$
Hence, $z = \cos  \displaystyle \frac{\pi}{8} + i  \sin  \frac{\pi}{8}$