Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $4\pm \dfrac {5}{4}b$
  2. $5\pm \dfrac {4}{5}a$
  3. $5\pm \dfrac {4}{5}b$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the ellipse x^2/9 + y^2/25 = 1, a^2=9, b^2=25. Since b > a, the foci are on the y-axis. e = sqrt(1 - 9/25) = 4/5. Foci are (0, +/- be) = (0, +/- 5 * 4/5) = (0, +/- 4). The distance from a point P(a,b) to the foci (0, 4) and (0, -4) is sqrt(a^2 + (b-4)^2) and sqrt(a^2 + (b+4)^2). Using the property of focal distances, this simplifies to 5 +/- (4/5)b.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $\infty$
  2. $4$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

An ellipse x^2/a^2 + y^2/b^2 = 1 with rational a^2 and b^2 has infinitely many rational points. This can be shown by parameterizing the ellipse using rational functions or by finding one rational point and using the chord method.

Multiple choice position of point wrt ellipse ellipse maths

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

  1. $\dfrac{6}{5}$
  2. $\dfrac{4}{5}$
  3. $\dfrac{3}{5}$
  4. $\dfrac{3}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that $2ae=6$ and $2b=8$

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

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

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

$16=a^2-9$

$\Rightarrow a^2=25$

$\Rightarrow a=5$

$5e=3$

$\Rightarrow e=\dfrac{3}{5}$.
Multiple choice position of point wrt ellipse ellipse maths

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

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

Multiple choice position of point wrt ellipse ellipse maths

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

  1. Outside the ellipse

  2. On the ellipse

  3. On the major axis

  4. On the minor axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice position of point wrt ellipse ellipse maths

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

  1. ($\sqrt{2}, \sqrt{2}$)
  2. ($0, \sqrt{3}$)
  3. ($\sqrt{5}, \dfrac{1}{\sqrt{2}}$)

  4. (2, 1)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

To find the point on the ellipse x^2 + 2y^2 = 6 closest to the line x + y = 7, the tangent to the ellipse must be parallel to the line x + y = 7 (slope = -1). The slope of the tangent to the ellipse is given by dy/dx = -x/(2y) = -1, which implies x = 2y. Substituting this into the ellipse equation gives (2y)^2 + 2y^2 = 6, leading to 6y^2 = 6, so y = 1 and x = 2. Thus, the coordinates are (2, 1).

Multiple choice position of point wrt ellipse ellipse maths

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

  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
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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.

Multiple choice position of point wrt ellipse ellipse maths

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

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

Multiple choice position of point wrt ellipse ellipse maths
$C: x^{2}+y^{2}=9$, $\displaystyle E: \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, $L: y=2x$

Let $L$ intersect $x=1$ at point $R$. Then which of the following is correct :
  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 $E$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $E$.

Multiple choice position of point wrt ellipse ellipse maths

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?

  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. $\dfrac{1}{2}+\dfrac{1}{\sqrt{e}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $(k, 2)$
$\therefore k=1+\dfrac{1}{2}-e^{-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{1}{\sqrt{e}}$.