Position of point wrt ellipse - class-XII

The region (disk) bounded by the ellipse is given by the equation:

The given ellipse equation is x^2 + 3y^2 = 6, which can be rewritten in standard form as x^2/6 + y^2/2 = 1. A point on the ellipse can be parameterized as (sqrt(6) cos(theta), sqrt(2) sin(theta)). The distance from the centre (0,0) is given as 2 units, so (sqrt(6) cos(theta))^2 + (sqrt(2) sin(theta))^2 = 2^2, which simplifies to 6 cos^2(theta) + 2 sin^2(theta) = 4. Solving this yields 4 cos^2(theta) + 2(cos^2(theta) + sin^2(theta)) = 4, so 4 cos^2(theta) + 2 = 4, meaning cos^2(theta) = 1/2. Thus cos(theta) = plus or minus 1/sqrt(2), making theta equal to pi/4, 3pi/4, etc., which are not in the options. Therefore, none of these is correct.

$f(x,y) = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - 1$ ........ $(1)$ 

The definition of an ellipse is the locus of a point whose distance from a focus is e times its distance from the directrix. Here e = 3/5 and the focus is (3,0). The directrix is x=p, so the equation is sqrt((x-3)^2 + y^2) = (3/5)|x-p|. Squaring both sides and simplifying to the form x^2/a^2 + y^2/b^2 = 1, we find the center is at (ae, 0). Since the center is at the origin, the focus must be at (ae, 0), implying ae = 3. With e = 3/5, a = 5. The directrix is x = a/e = 5/(3/5) = 25/3.

The point $(x,y)$ lies inside the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ if it satisfies the following condition i.e.

The points (3,0), (0,4), and (4,0) do not satisfy the standard ellipse equation x^2/a^2 + y^2/b^2 = 1 unless specific values for a and b are given. Given the context of standard coordinate geometry problems, these points do not lie on a single ellipse centered at the origin with these intercepts.

Given equation is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } =1$

$ \dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1=0$

The curve is defined by, 

$f(x,y)=\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { { (2) }^{ 2 } }{ 16 } +\dfrac { { (3) }^{ 2 } }{ 25 } -1\\ f(2,3)=\dfrac { 1 }{ 4 } +\dfrac { 9 }{ 25 } -1=\dfrac { 25+36-100 }{ 100 } =-\dfrac { 39 }{ 100 } \\ f(2,3)<0$

So, the point lies inside the ellipse.

Option B is correct.

Given equation is $16{ x }^{ 2 }+9{ y }^{ 2 }=144$

Given ellipse $\dfrac {x^2}{4}+\dfrac {y^2}{3}=1$

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

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.

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.

Equation of ellipse-

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.

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).

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.

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 $cos^2\theta+sin^2\theta=1$)

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 = (20 + 1x_B)/3 and y = (2y_A + 10)/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.

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.

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.

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

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).

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.

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) $

$y=2x$, 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 $E$.

Since $\displaystyle \frac { { 1 }^{ 2 } }{ 9 } +\frac { { 2 }^{ 2 } }{ 4 } -=\frac{1}{9}>0$

Let the required equation of the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.

$Equation\quad of\quad ellipse:\quad 4{ x }^{ 2 }+5{ y }^{ 2 }=1\ Putting\quad the\quad point\quad (4,-3)\quad on\quad the\quad ellipse\quad we\quad get:\ \quad =4{ (4) }^{ 2 }+5{ (-3) }^{ 2 }-1\ =\quad 64+45-1=28>0\ \therefore \quad The\quad point\quad lies\quad outside\quad the\quad ellipse.$

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

$24x^{2}+7y^{2}-12=5$

$S(5,-2)=24\times 25+7\times 4-12$

$S(5,-2)>0$

$\therefore $ lies outside the ellipse.
$S=0$
$S(x _{1}y _{1})>0$ the point is outside the ellipse.

Given point is $\left ( 2\cos\theta, 3\sin\theta  \right )$

Equation of ellipse is $\dfrac { { x }^{ 2 } }{ 16 } +\dfrac { { y }^{ 2 } }{ 9 } =1$

Ellipse : $16{ x }^{ 2 }+9{ y }^{ 2 }-16x-32=0$

We will substitute the given point in ellipse equation

$P(\sqrt { 6 } \cos { \theta  } ,\sqrt { 2 } \sin { \theta  } )$

Equation of ellipse,  $\dfrac { { x }^{ 2 } }{ 9 } +\dfrac { { y }^{ 2 } }{ 16 } =1$

Equation of ellipse : $\dfrac { { \left( x-3 \right)  }^{ 2 } }{ 9 } +\dfrac { { \left( y-4 \right)  }^{ 2 } }{ 16 } -1=0$

Equation of ellipse :   ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

Equation of ellipse : ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

If equation of ellipse is $\dfrac { { \left( x-h \right)  }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { \left( y-k \right)  }^{ 2 } }{ { b }^{ 2 } } =1$,  centre $\left( h,k \right) $

Equation of ellipse    ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$

          ${ 5x }^{ 2 }+{ 7y }^{ 2 }=140$          (equation of ellipse)

$f\left( x,y \right) ={ 2x }^{ 2 }+7{ y }^{ 2 }-20$

Point will be outside of ellipse if $f\left( 1,2 \right) >0$

on ellipse if $f\left( 1,2 \right) =0$

will be inside if $f\left( 1,2 \right) <0$

$ f\left( 1,2 \right) ={ 2\times 1 }^{ 2 }+7{ \times 2 }^{ 2 }-20=10$

Here we see that $f\left( 1,2 \right) >0$ so point will be outside of ellipse

So correct answer will be option A

The ellipse is x^2/(4c^2) + y^2/c^2 = 1. The circle is x^2 + y^2 = 9a^2. For four distinct intersection points, the circle must be larger than the minor axis but smaller than the major axis of the ellipse. This leads to the inequality 9ac - 9a^2 - 2c^2 < 0.

The area of the circle is pia^2. The area of the ellipse is pia*b. Probability of being outside = 1 - (Area_ellipse/Area_circle) = 1 - b/a = 2/3. Thus b/a = 1/3. Since b^2 = a^2(1-e^2), we have (b/a)^2 = 1-e^2. So 1/9 = 1-e^2, e^2 = 8/9, e = sqrt(8)/3 = 2sqrt(2)/3.

Equation of tangent at $P (a cos \theta, b sin \theta)$ is
$\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1$
$\therefore$ Equation of the lines OA and OB is
$x^{2}+y^{2}-a^{2}\left(\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}\right)^{2}=0$
since the lines OA and OB are perpendicular
$\therefore$ $1- \cos ^{2}\theta + 1 - \frac{a^{2}}{b^{2}} \sin^{2}\theta=0$
i.e $\sin^{2}\theta + 1-\frac{a^{2}}{b^{2}} \sin^{2}\theta$
i.e $\frac{b^{2}}{a^{2}}=\frac{sin^{2}\theta}{1+ \sin^{2} \theta}$
$\therefore e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{\sin^{2} \theta}{1+\sin^{2}\theta}=\frac{1}{1+\sin^{2}\theta}$
$\therefore$ e is least if $\theta = \pm \frac{\pi}{2}$
$\therefore$ the point P is $(0,\pm b)$

The center of ellipse is $(0,0)$

$4x^2+9y^2=36\Rightarrow \dfrac {x^2}9+\dfrac {y^2}4=1$

• Substitute each and every point on given ellipse and see which will satisfy the equation
• Take point $(1.4,3.2)$ , when we substitute we get $4(1.96)+9(10.24) = 100$
• Therefore correct answer is option $B$