Tag: ellipse

Questions Related to ellipse

The point at shortest distance from the line x+y=7 and lying on an ellipse $x^2 + 2y^2 =6$, has coordinates

position of point wrt ellipse ellipse maths

  1. ($\sqrt{2}, \sqrt{2}$)

  2. ($0, \sqrt{3}$)

  3. ($\sqrt{5}, \dfrac{1}{\sqrt{2}}$)

  4. (2, 1)

Show answer
Correct Option: A

Which of the following points is an exterior point of the ellipse $\displaystyle 16 x^{2} + 9y^{2} - 16x - 32 = 0$.

position of point wrt ellipse ellipse maths

  1. $\displaystyle \left ( \frac{1}{2}, : 2 \right )$

  2. $\displaystyle \left ( \frac{1}{4}, : 2 \right )$

  3. $\displaystyle \left ( 3, : 2 \right )$

  4. none of these

Show answer
Correct Option: B,C
Explanation:

Let   $S = \displaystyle 16 x^{2} + 9y^{2} - 16x - 32 $
Now $S(\dfrac12,2)=4+36-8-32 = 0 \Rightarrow $ point on the ellipse.
$S(\dfrac14,2) = 1+36-4-32> 0 \Rightarrow $ point is exterior to the ellipse.
$S(3,2) = 144+36-48-32>0 \Rightarrow $ point is exterior to the ellipse.

An ellipse with foci $(0,\pm 2)$ has length of minor axis as $4$ units. Then the ellipse will pass through the point

position of point wrt ellipse ellipse maths

  1. $\left( 2,\sqrt { 2 } \right) $

  2. $\left( \sqrt { 2 } ,2 \right) $

  3. $\left( 2,2\sqrt { 2 } \right) $

  4. $\left( 2\sqrt { 2 } ,2 \right) $

Show answer
Correct Option: B
Explanation:

Let $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1(a<b)\quad $ is the equation of ellipse, foci $(0,\pm 2)$
(be $=2$)
Given: $2a=4\Rightarrow a=2$
${ e }^{ 2 }=1-\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } \Rightarrow { b }^{ 2 }{ e }^{ 2 }={ b }^{ 2 }-{ a }^{ 2 }\quad $
$\quad 4={ b }^{ 2 }-4\Rightarrow { b }^{ 2 }=8$
$\therefore$ equation of ellipse is $\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { y }^{ 2 } }{ 8 } =1\quad $
It passes through $\left( \sqrt { 2 } ,2 \right) $

$C: x^{2}+y^{2}=9$, $\displaystyle E: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, $L: y=2x$

Let $L$ intersect $x=1$ at point $R$. Then which of the following is correct :
position of point wrt ellipse ellipse maths

  1. $R$ lies inside both $C$ and $E$

  2. $R$ lies outside both $C$ and $E$

  3. $R$ lies on both $C$ and $E$

  4. $R$ lies inside $C$ but outside $E$

Show answer
Correct Option: D
Explanation:

$y=2x$, intersects $x=1$ at $(1,2)$
Coordinate of $R$ are $(1,2)$
$C(1,2)=1+22-9<0$ Since $C(1,2)$ is $<0, R $ lies inside $C$
$E(1,2)=\dfrac{1}9+1-1>0$ Since $E(1,2)$ is $>0, R $ lies outside $E$.

Let a curve satisfying the differential equation $y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$ which passes through $(1, 1)$. If the curve also passes through $(k, 2)$, then value of k is?

position of point wrt ellipse ellipse maths

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

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

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

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

Show answer
Correct Option: C
Explanation:

$y^2dx+\left(x-\dfrac{1}{y}\right)dy=0$
$\Rightarrow \dfrac{dx}{dy}+\dfrac{x}{y^2}=\dfrac{1}{y^3}$
Integrating factor (I.F.)$=e^{-\dfrac{1}{y}}$
Now $x.e^{-\dfrac{1}{y}}=\displaystyle\int e^{-\dfrac{1}{y}}\dfrac{1}{y^3}dy$
Put $-\dfrac{1}{y}=y$
$x.e^t=\displaystyle\int e^t(-t)dt$
$\Rightarrow x.e^t=-(t.e^t-e^t)+c$
$\Rightarrow e^{-\dfrac{1}{y}}=e^{-\dfrac{1}{y}}\left(1+\dfrac{1}{y}\right)+c$
$\Rightarrow x=1+\dfrac{1}{y}+c.e^{\dfrac{1}{y}}$
it passes through point $(1, 1)$
$\therefore c=-\dfrac{1}{e}$
Equation of curve is
$x=1+\dfrac{1}{y}-e^{\dfrac{1}{y}-1}$
It passes through $(k, 2)$
$\therefore k=1+\dfrac{1}{2}-e^{-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}$.

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

position of point wrt ellipse ellipse maths

  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 $E$

Show answer
Correct Option: D
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 $E$.

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

position of point wrt ellipse ellipse maths

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

Show answer
Correct Option: A
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 $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$.

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

position of point wrt ellipse ellipse maths

  1. lies on the curve

  2. lies inside the curve

  3. lies outside the curve

  4. lies focus of the curve

Show answer
Correct Option: C
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]

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

position of point wrt ellipse ellipse maths

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

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

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

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

Show answer
Correct Option: A
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 $\theta =\dfrac{\pi }{4}$ 

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

position of point wrt ellipse ellipse maths

  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

Show answer
Correct Option: B
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) $