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Questions Related to maths

Multiple choice position of point wrt ellipse ellipse 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

  1. $8$
  2. $6$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice position of point wrt ellipse ellipse maths

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

  1. outside the ellipse

  2. inside the ellipse but not at the focus

  3. on the ellipse

  4. at the focus
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice position of point wrt ellipse ellipse maths

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

  1. a-b

  2. a+b

  3. 2a+2b

  4. 2(a-b)
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $a+b$

Multiple choice position of point wrt ellipse ellipse maths

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

  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. $9ac-9{ a }^{ 2 }-2{ c }^{ 2 }>0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ellipse is x^2/(4c^2) + y^2/c^2 = 1. The circle is x^2 + y^2 = 9a^2. For four distinct intersection points, the circle must be larger than the minor axis but smaller than the major axis of the ellipse. This leads to the inequality 9ac - 9a^2 - 2c^2 < 0.

Multiple choice position of point wrt ellipse ellipse maths

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:

  1. $\dfrac{2\sqrt{2}}{3}$
  2. $\dfrac{\sqrt{5}}{3}$
  3. $\dfrac{8}{9}$
  4. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The area of the circle is pi*a^2. The area of the ellipse is pi*a*b. Probability of being outside = 1 - (Area_ellipse/Area_circle) = 1 - b/a = 2/3. Thus b/a = 1/3. Since b^2 = a^2(1-e^2), we have (b/a)^2 = 1-e^2. So 1/9 = 1-e^2, e^2 = 8/9, e = sqrt(8)/3 = 2sqrt(2)/3.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $(0,ae)$
  2. $(a,0)$
  3. $(-a,0)$
  4. $(0,-b)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(0,\pm b)$

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $(\sqrt 3, 2)$
  2. $(-\sqrt 3, 2)$
  3. $(-\sqrt 3, -2)$
  4. $(\sqrt 3, -2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice position of point wrt ellipse ellipse maths

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

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

  4. NONE
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $(1, 3.5)$
  2. $(1.4, 3.2)$
  3. $(1.9, 2.9)$
  4. $(2.3, 3.1)$
  5. $(2.7, 2.6)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $B$
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

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

  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. $i$
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $z = \cos  \displaystyle \frac{\pi}{8} + i  \sin  \frac{\pi}{8}$