Tag: maths

Questions Related to maths

The distance from the foci of $P(a,b)$ on the ellipse $\dfrac {x^{2}}{9}+\dfrac {y^{2}}{25}=1$ are

position of point wrt ellipse ellipse maths

  1. $4\pm \dfrac {5}{4}b$

  2. $5\pm \dfrac {4}{5}a$

  3. $5\pm \dfrac {4}{5}b$

  4. $none\ of\ these$

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Correct Option: C

The number of rational points on the ellipse $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$ is

position of point wrt ellipse ellipse maths

  1. $\infty$

  2. $4$

  3. $0$

  4. $2$

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Correct Option: A

In an ellipse the distance between its foci is 6 and its minor axis is 8 . Its eccentricity is

position of point wrt ellipse ellipse maths

  1. $\dfrac{6}{5}$

  2. $\dfrac{4}{5}$

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

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

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Correct Option: C
Explanation:
Given that $2ae=6$ and $2b=8$

$\Rightarrow ae=3$      $\Rightarrow b=4$

$b^2=a^2(1-e^2)$

$b^2=a^2-a^2e^2$

$16=a^2-9$

$\Rightarrow a^2=25$

$\Rightarrow a=5$

$5e=3$

$\Rightarrow e=\dfrac{3}{5}$.

A point on the ellipse is $\displaystyle \frac{x^{2}}{6} + \frac{y^{2}}{2} = 1$ at a distance of $2$ from the centre of the ellipse has the eccentric angle

position of point wrt ellipse ellipse maths

  1. $\displaystyle \frac{\pi}{4}$

  2. $\displaystyle \frac{\pi}{3}$

  3. $\displaystyle \frac{\pi}{6}$

  4. $\displaystyle \frac{\pi}{2}$

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Correct Option: A
Explanation:

Given, equation of ellipse as $\displaystyle\frac{x^2}{6}+\displaystyle\frac{y^2}{2}=1$(where length of major axis=$\sqrt6$,length of minor axis=$\sqrt2$)
center of ellipse is (0,0) which is parallel to horizontal axis and with eccentricity 'e'.
Any point on the ellipse will be as $(acos\theta,bsin\theta)\Rightarrow P(\sqrt6cos\theta,\sqrt2sin\theta)$
Distance of point P from center=2
$\Rightarrow \sqrt((\sqrt6cos\theta-0)^2+(\sqrt2sin\theta-0)^2)=2$
$\Rightarrow (6cos^2\theta+2sin^2\theta)=4$
$\Rightarrow (3cos^2\theta+sin^2\theta)=2$
$\Rightarrow 2cos^2\theta+1=2$
$\Rightarrow cos\theta=\pm\frac{1}{\sqrt2}$
$\Rightarrow \theta=\displaystyle\frac{\pi}{4}\;or\;\displaystyle\frac{-\pi}{4}$
Option $A$ is correct

The position of the point $(1, 3)$ with respect to the ellipse $4x^2+9y^2-16x-54y+61=0$.

position of point wrt ellipse ellipse maths

  1. Outside the ellipse

  2. On the ellipse

  3. On the major axis

  4. On the minor axis

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Correct Option: A

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)

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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

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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 $E$

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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 $(k, 2)$
$\therefore k=1+\dfrac{1}{2}-e^{-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}$.