Tag: ellipse

Questions Related to ellipse

The segment of the tangent at the point P to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, intercepted by the auxiliary circle subtends a right angle at the origin. If the eccentricity of the ellipse is smallest possible, then the point P can be

position of point wrt ellipse ellipse maths

  1. $(0,ae)$

  2. $(a,0)$

  3. $(-a,0)$

  4. $(0,-b)$

Show answer
Correct Option: D
Explanation:

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 $(0,\pm b)$

If one end of the diameter of the ellipse $4x^2+y^2=16$ is $(\sqrt 3, 2)$, then the other end is:

position of point wrt ellipse ellipse maths

  1. $(\sqrt 3, 2)$

  2. $(-\sqrt 3, 2)$

  3. $(-\sqrt 3, -2)$

  4. $(\sqrt 3, -2)$

Show answer
Correct Option: C
Explanation:

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

The diameter of ellipse will pass through the center of ellipse and center will divide the diameter into two halfs
One end of diameter is $(\sqrt3,2) $ , Let the other end be $(h,k)$
We have $h+\sqrt3=0$ and $k+2=0$
$\Rightarrow h=-\sqrt3,k=-2$
Therefore option $C$ is correct

The point P on the ellipse $4x^2+9y^2=36$ is such that the area of the $\Delta PF _1F _2=\sqrt{10} Sq$ units, where $F _1.F _2$ are Foci. Then P has the coordinates

position of point wrt ellipse ellipse maths

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

  2. $(\dfrac{3}{2},2)$

  3. $(\dfrac{-3}{2},-2)$

  4. NONE

Show answer
Correct Option: A
Explanation:

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

So, $a=3,b=2$

In an ellipse, Focal points, $F=(\pm\sqrt {a^2-b^2},0)=(\pm\sqrt 5,0)$
So, $F _1F _2=2\sqrt 5$ which is the base of the triangle $PF _1F _2$

An arbitrry point P on ellipse is $(acos\theta,b\sin\theta)$
So, $Height\ of\ \triangle PF _1F _2=b\sin\theta=2\sin\theta$

ie, $Area=\dfrac 12base\times height=\dfrac 12\times 2\sqrt 5\times2\sin\theta=2\sqrt 5\sin\theta=\sqrt {10}$
$\Rightarrow \sin\theta=\dfrac 1{\sqrt 2}$
$\Rightarrow \cos\theta=\pm\dfrac 1{\sqrt 2}$
And $P=(a\cos\theta,b\sin\theta)=(\pm\dfrac 3{\sqrt 2},\sqrt 2)$

So, Option$ A$ has one of the anwers.

Which of the following is an (x,y) coordinate pair located on the ellipse $4x^2 + 9y^2 = 100$?

position of point wrt ellipse ellipse maths

  1. $(1, 3.5)$

  2. $(1.4, 3.2)$

  3. $(1.9, 2.9)$

  4. $(2.3, 3.1)$

  5. $(2.7, 2.6)$

Show answer
Correct Option: B
Explanation:
  • 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 $B$