Tag: position of point wrt ellipse

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Questions Related to position of point wrt ellipse

Multiple choice position of point wrt ellipse ellipse maths

Let $(a, 0)$ and $B(b, 0)$ be fixed distinct points on the $x-axis$, none of which coincides with the origin $O(0, 0)$ and let $C$ be a point on the $y-axis$. Let $L$ be a line through the $O(0, 0)$ and perpendicular to the line $AC$, The locus of the point of intersection of lines $L$ and $BC$ if $C$ varies along the $y-axis$, is (provided $x^{2}+ab\neq 0$) 

  1. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=x$
  2. $\dfrac{x^{2}}{a}+\dfrac{y^{2}}{b}=y$
  3. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=x$
  4. $\dfrac{x^{2}}{b}+\dfrac{y^{2}}{a}=y$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let C = (0, c). A = (a, 0), B = (b, 0). Line AC: y - 0 = (c-0)/(0-a) * (x-a) => y = -c/a * (x-a). Line L is perpendicular to AC through origin: y = a/c * x. Line BC: y - 0 = (c-0)/(0-b) * (x-b) => y = -c/b * (x-b). Solving for the intersection of L and BC by eliminating c, we get x^2/b + y^2/a = y.

Multiple choice position of point wrt ellipse ellipse maths

If P($\theta$) and Q($\pi$/2 + $\theta$) are two points on the ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Locus of the mid-point of PQ is

  1. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{2}$
  2. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 4$
  3. $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,$P(\theta),Q(\frac{\pi}{2}+\theta)$ are two points on the ellipse $\displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=1$
Any point on the ellipse will be $(acos\theta,bsin\theta)$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(acos(\frac{\pi}{2}+\theta),bsin(\frac{\pi}{2}+\theta))$
$\Rightarrow P=(acos\theta,bsin\theta),Q=(-asin\theta,bcos\theta)$
Let required point be $C(x,y)$
Given, $C=mid-point\;of\;PQ$
$\Rightarrow (x,y)=(\displaystyle\frac{(acos\theta-asin\theta)}{2},\displaystyle\frac{(bsin\theta+bcos\theta)}{2})$
$\Rightarrow \displaystyle\frac{x}{a}=(\displaystyle\frac{(cos\theta-sin\theta)}{2}),\displaystyle\frac{y}{b}=(\displaystyle\frac{(sin\theta+cos\theta)}{2})$
on squaring $\displaystyle\frac{x}{a},\displaystyle\frac{y}{b}$ and adding both
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=(\displaystyle\frac{(cos\theta-sin\theta)}{2})^2+(\displaystyle\frac{(sin\theta+cos\theta)}{2})^2$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{2(cos^2\theta+sin^2\theta)}{4}$
$\Rightarrow \displaystyle\frac{x^2}{a^2}+\displaystyle\frac{y^2}{b^2}=\displaystyle\frac{1}{2}$(since $cos^2\theta+sin^2\theta=1$)

Multiple choice position of point wrt ellipse ellipse maths

The value of $\alpha$ for which the point $(\alpha,\alpha+2)$ is an interior point of smaller segment of the curve $x^{2}+y^{2}-4=0$ made by the chord of the curve whose equation is $3x+4y+12=0$ is

  1. $\left(-\infty,\dfrac {-20}{7}\right)$
  2. $(-2,0)$
  3. $\left(-\infty,\dfrac {20}{7}\right)$
  4. $\alpha\ \epsilon\ \phi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice position of point wrt ellipse ellipse maths

The distance of a point on the ellipse $\dfrac {x^{2}}{6}+\dfrac {y^{2}}{2}=1$ from the centre is $2$, then the eccentric angle is-

  1. $\dfrac \pi3$
  2. $\dfrac \pi4$
  3. $\dfrac \pi6$
  4. $\dfrac \pi2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given ellipse $ \dfrac{x^{2}}{6}+\dfrac{y^{2}}{2} = 1 $

Let $ \theta$ be the eccentric angle of the point $p$

coordinate of $p$ $ (\sqrt{6}cos\theta ,\sqrt{2}sin\theta )$

Given distance $= 2 $

$ \therefore $ $OP = 2$

$ \sqrt{6 cos^{2}\theta +2sin^{2}\theta } = 2 \Rightarrow 6cos^{2}\theta +2sin^{2}\theta  = 4$

$ 3cos^{2}\theta +sin^{2}\theta  = 2 $

$ 2 sin^{2}\theta  = 1$

$ sin^{2}\theta  = \dfrac{1}{2} \Rightarrow  sin\theta  = \pm  \dfrac{1}{\sqrt{2}}$

$ \therefore $ eccentric angle $\theta  = \pm \dfrac{\pi }{4}$
Multiple choice position of point wrt ellipse ellipse maths

A rod of length $l$ rests against a vertical wall and a floor of a room.Let P be a point on the rod,nearer to its end on the wall, that divides its length in the ratio 1:2 if the rod begins to slide on the floor,then the locus of P is:

  1. an ellipse of eccentricity $\dfrac { 1 }{ 2 }$
  2. an ellipse of eccentricity $\dfrac { \sqrt { 3 } }{ 2 }$
  3. a circle of radius $\dfrac { l }{ 2 }$
  4. a circle of radius $\dfrac { \sqrt { 3 } }{ 2 } l$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the rod be AB of length l, with A on the y-axis and B on the x-axis. A point P(x,y) dividing the rod in ratio 1:2 has coordinates x = (2*0 + 1*x_B)/3 and y = (2*y_A + 1*0)/3. Thus x = x_B/3 and y = 2y_A/3. Since x_B^2 + y_A^2 = l^2, we have (3x)^2 + (3y/2)^2 = l^2, which is 9x^2 + 9y^2/4 = l^2. This is an ellipse with semi-axes a = l/3 and b = 2l/3. Eccentricity e = sqrt(1 - (l/3)^2/(2l/3)^2) = sqrt(1 - 1/4) = sqrt(3)/2.

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