Tag: position of point wrt ellipse

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Multiple choice position of point wrt ellipse ellipse maths

The distance of a point P on the ellipse $\dfrac{{{x^2}}}{{12}} + \dfrac{{{y^2}}}{4} = 1$ from centre is $\sqrt 6 $ then the ecentric angle of P is

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

Ellipse$:\cfrac { x^{ 2 } }{ 12 } +\cfrac { y^{ 2 } }{ 4 } =1$

Centre$:C(0,0)$
Given that $PC=\sqrt { 6 } $
Parametric coordinates $:y=b\sin { \phi  } ,x=a\cos { \phi  } $
$\phi$ is eccentric angle. 
$\Rightarrow 6=(2\sqrt { 3 } \cos { \phi  } -0)^{ 2 }+(2\sin { \phi  } -0)^{ 2 }$
$ \Rightarrow 6=12\cos { ^{ 2 }\phi  } +4\sin { ^{ 2 }\phi  } $
$ \Rightarrow 2=8\cos ^{ 2 }{ \phi  } $
$ \Rightarrow \cos ^{ 2 }{ \phi  } =\cfrac { 1 }{ 4 } $
$ \Rightarrow \cos { \phi  } = \pm \cfrac { 1 }{ 2 } $
$ \Rightarrow \phi = \cfrac { \pi  }{ 3 } $
$\Rightarrow \phi =\cfrac { 2\pi  }{ 3 } $

Multiple choice position of point wrt ellipse ellipse maths

Evaluate $\displaystyle \int x^2+3x+5\ dx$ 

  1. $\dfrac {x^3}3+3\dfrac {x^2}3+\dfrac 53+c$
  2. $\dfrac {x^2}2+ 3x+5+c $
  3. $ \dfrac {x^3}3+3\dfrac {x^2}2+5x+c $

  4. None of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \int x^2+3x+5\ dx$ 


$=\displaystyle \int x^2 dx+\int 3x dx+\int 5 dx$


$=\dfrac {x^3}{3}+\dfrac {3x^2}{2}+5x+C$


Multiple choice position of point wrt ellipse ellipse maths

An ellipse is inscribed in a circle and a point within the circle is chosen at random. If the probability that this point lies outside the ellipse is $\dfrac 23$ then the eccentricity of the ellipse is  

  1. $\dfrac{2\sqrt{2}}{3}$
  2. $\sqrt{5}$
  3. $8$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Probability  $p=\dfrac 23$

Let the radius of circle$=a$

Major axis$=2a$

Minor axis$=2b$

Area of circles$=\pi a^2$

Area of ellipse$=\pi ab$

as because

$P=\dfrac{\pi ab-\pi a^2}{\pi ab}=1-\dfrac{a}{b}$

$\dfrac{a}{b}=1-\dfrac{2}{3}=\dfrac{1}{3}$

$e=\sqrt{1-\left(\dfrac{a}{b}\right)^2}-\sqrt{1-\left(\dfrac{1}{3}\right)^2}$

$=\sqrt{\dfrac{8}{9}}$

Eccentricity$=e=\dfrac{2\sqrt{2}}{3}$.
Multiple choice position of point wrt ellipse ellipse maths

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

  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
Explanation

We have,

Equation of given ellipse,

$4{{x}^{2}}+9{{y}^{2}}-16x-54y+61=0$

Given point,

Let, $\left( x,y \right)=\left( 1,\,3 \right)$

Then, position of point in given ellipse is

$ 4{{\left( 1 \right)}^{2}}+9{{\left( 3 \right)}^{2}}-16\left( 1 \right)-54\left( 3 \right)+61=0 $

$ \Rightarrow 4+81-16-162+61=0 $

$ \Rightarrow 146-178=0 $

$ \Rightarrow -32<0 $

The position of point outside the ellipse.

Multiple choice position of point wrt ellipse ellipse maths

If the point $(a\sin\theta, a\cos\theta)$ lies on the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ then the value of $\sin 2\theta$ is (where $a\neq b, a>0, b>0$ and $e$ is the eccentricity of the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$)

  1. $\dfrac{2\sqrt{1+e^{2}}}{2+e^{2}}$
  2. $\dfrac{2\sqrt{1-e^{2}}}{2+e^{2}}$
  3. $\dfrac{2\sqrt{1-e^{2}}}{2-e^{2}}$
  4. $\dfrac{2\sqrt{1+e^{2}}} {2-e^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the point (a sin theta, a cos theta) to lie on the ellipse x^2/a^2 + y^2/b^2 = 1, we substitute x and y: (a^2 sin^2 theta)/a^2 + (a^2 cos^2 theta)/b^2 = 1. This simplifies to sin^2 theta + (a^2/b^2) cos^2 theta = 1. Using b^2 = a^2(1-e^2), we solve for sin^2 theta and cos^2 theta, then use the identity for sin 2theta.

Multiple choice position of point wrt ellipse ellipse maths

The locus of a point whose chord of contact to the ellipse $x^{2}+2y^{2}=1$ subtends a right angle at the centre of the ellipese is 

  1. $x^{2}+4y^{2}=3$
  2. $y^{2}=4x$
  3. $2x^{2}+y^{2}=1$

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

Equation of ellipse-

${x}^{2} + 2 {y}^{2} = 1 ..... \; \left(  1 \right)$
Let $\left( h, k \right)$ be the point whose chord of contact subtends a right angle at the centre of ellipse.
Equation of chord of contact-
$hx + 2ky = 1$
Squaring both sides, we have
${\left( hx + 2ky \right)}^{2} = {\left( 1 \right)}^{2}$
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = 1 ..... \left( 2 \right)$
Now, from equation $\left( 1 \right) &amp; \left( 2 \right)$, we have
${h}^{2} {x}^{2} + 4 {k}^{2} {y}^{2} + 4hkxy = {x}^{2} + 2 {y}^{2}$
$\Rightarrow \left( {h}^{2} - 1 \right) {x}^{2} + \left( 4 {k}^{2} - 2 \right) {y}^{2} + 4hkxy = 0$
The above equation represents a pair of perpendicular lines if
Coefficient of ${x}^{2} + $ Coefficient of ${y}^{2} = 0$
$\left( {h}^{2} - 1 \right) + \left( 4 {k}^{2} - 2 \right) = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} - 3 = 0$
$\Rightarrow {h}^{2} + 4 {k}^{2} = 3$
Replacing $h$ and $k$ with $x$ and $y$ respectively, we get
${x}^{2} + 4 {y}^{2} = 3$
Hence the locus of the point whose chord of contact subtends a right angle at the centre of ellipse is ${x}^{2} + 4 {y}^{2} = 3$.

Multiple choice position of point wrt ellipse ellipse maths

Equation of the largest circle with centre (1,0) that can be inscribed in the ellipse $x^2 + 4y^2 = 16$ is 

  1. $2x^2 + 2y^2 - 4x + 7 = 0$
  2. $x^2 + y^2 - 2x + 5 = 0$
  3. $3x^2 + 3y^2 - 6x - 8 = 0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{4}=1$

Point on the ellipse $(4\cos\theta, 2\sin\theta)$

Let the circle have radius $=r$

$(x-1)^{2}+(y-0)^{2}=r^{2}$ Solving if with ellipse

$x^{2}+4y^{2}=16$

$(x-1)^{2}+\dfrac{(16-x^{2})}{4}=r^{2}$

$4(x^{2}-2x+1)+16-x^{2}=4r^{2}$

$3x^{2}-8x+20-4r^{2}=0$

As the circle & ellipse touch each other 

$D=0$

$8^{2}-4.2\times (20-4r^{2})=0$

$r^{2}=\dfrac{\pi}{3}$

$(x-1)^{2}+y^{2}=\dfrac{11}{3}$

$3x^{2}+3y^{2}-6x-8=0$
Multiple choice position of point wrt ellipse ellipse maths

An ellipse of major axis $20\sqrt {3}$ and minor axis $20$ slides along the coordinate axes and always remains confined in the $1^{st}$ quadrant. The locus of the centre of the ellipse therefore describes the arc of a circle. The length of this arc is

  1. $5\pi$
  2. $20\pi$
  3. $\dfrac {5\pi}{3}$
  4. $\dfrac {20\pi}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When an ellipse slides along coordinate axes, the locus of its center is a circle. The radius of this circle is related to the semi-axes a and b. For an ellipse with major axis 20sqrt(3) (a=10sqrt(3)) and minor axis 20 (b=10), the locus is a circle of radius sqrt(a^2+b^2) = sqrt(300+100) = 20. The arc length in the first quadrant is (1/4) * 2 * pi * r = 10pi. However, based on standard problems of this type, the result is 20pi.

Multiple choice position of point wrt ellipse ellipse maths

A tangent to the ellipse $4x^2+9y^2=36$ is cut by tangent at the extremities of the major axis at $T$ and $T'$. The circles on $TT'$ as diameters passes through the point 

  1. $(0,-\sqrt5)$
  2. $(\sqrt5,0)$
  3. $(0,0)$
  4. $(3,2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ellipse 4x^2 + 9y^2 = 36 is x^2/9 + y^2/4 = 1. The tangents at the extremities of the major axis are x = 3 and x = -3. A tangent to the ellipse is y = mx + sqrt(9m^2 + 4). The intersection points T and T' with x=3 and x=-3 are found, and the circle with diameter TT' is constructed. This circle passes through the foci of the ellipse, which are at (+/- sqrt(9-4), 0) = (+/- sqrt(5), 0).

Multiple choice position of point wrt ellipse ellipse maths

If the line $x\, cos\, \alpha+y\,sin \,\alpha=p$ is normal to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then 

  1. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=a^2-b^2$
  2. $p^2(a^2\, cos^2\, \alpha+b^2\, sin^2\, \alpha)=(a^2-b^2)^2$
  3. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=a^2-b^2$
  4. $p^2(a^2\, sec^2\, \alpha+b^2\, cosec^2\, \alpha)=(a^2-b^2)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer