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Questions Related to maths

Multiple choice position of point wrt ellipse ellipse 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

  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 $E$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $E$.

Multiple choice position of point wrt ellipse ellipse maths

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

  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. $\dfrac{x^2}{9}+\dfrac{y^2}{36}=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. lies on the curve

  2. lies inside the curve

  3. lies outside the curve

  4. lies focus of the curve
Reveal answer Fill a bubble to check yourself
C Correct answer
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]

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $\dfrac{\pi }{4}$
  2. $\dfrac{\pi }{2}$
  3. $\dfrac{\pi }{6}$
  4. $\dfrac{\pi }{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\theta =\dfrac{\pi }{4}$ 

Multiple choice position of point wrt ellipse ellipse maths

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

Multiple choice position of point wrt ellipse ellipse maths

$\mathrm{A}$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 $S _{11}>0$ 

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. outside

  2. inside

  3. on the periphery of

  4. on the auxillary of
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $a(e + \cos \theta)$
  2. $a(e - \cos \theta)$
  3. $a(1 + e \cos \theta)$
  4. $a(1 + 2e \cos \theta)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $FP=\sqrt{(-ae-cos\theta)^2+(bsin\theta-0)^2}$
$=a \sqrt{(e^2+cos^2 \theta+2ecos\theta+(1-e^2)sin \theta-0)^2}$
$a\sqrt{(e^2+cos^2 \theta+2ecos\theta+sin^2 \theta-e^2sin^2\theta)}$
$a\sqrt{(1+2ecos\theta+e^2(1-sin^2\theta)}$
$a\sqrt{(1+2ecos \theta+e^2cos^2\theta)}$
$a\sqrt{(1+ecos\theta)^2}$
$FP=a(1+ecos\theta)$



Multiple choice position of point wrt ellipse ellipse maths

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

  1. Inside

  2. Outside

  3. On

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice position of point wrt ellipse ellipse maths

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

  1. outside

  2. inside

  3. on the periphery

  4. on the auxiliary circle
Reveal answer Fill a bubble to check yourself
D Correct answer
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.