Tag: equation of ellipse

Questions Related to equation of ellipse

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If S and S' are the foci of an ellipse of major axis of length 10 units and P is any point on the ellipse such that the perimeter of triangle PSS' is 15 units, then the eccentricity of the ellipse is 

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$
  3. $\dfrac{7}{25}$
  4. $\dfrac{3}{4}$
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Explanation

Perimeter = PF1 + PF2 + F1F2 = 2a + 2ae = 15. Given 2a = 10, then 10 + 10e = 15, so 10e = 5, e = 1/2.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If normal at any point P on the ellipse $\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1(a>b>0)$ meet the major and minor axes at Q and R respectively so that 3PQ = @PR, then the eccentricity of ellipse is equal to

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

Using the normal equation at P(a cos theta, b sin theta) and the ratio of segments PQ and PR, one can solve for the eccentricity.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

Find the length of the semi-axes, coordinates of foci, length of latus rectum, eccentricity and equation direction for the ellipse given by the equations :-  (i) $25{ x }^{ 2 }-150x+16{ y }^{ 2 }=175$ (ii) The eccentricity of the ellipse $9{ x }^{ 2 }+4{ y }^{ 2 }30y=0$ is 

  1. 1/2

  2. 2/3

  3. 3/4

  4. None of these

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Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If normal to the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ at $\left(ae,\dfrac{b^{2}}{a}\right)$ is passing throught $\left(0,-2b\right)$, then $c=$

  1. $\dfrac{1}{2}$
  2. $2\left(\sqrt{2}-1\right)$
  3. $\sqrt{2\sqrt{2}-2}$
  4. $\dfrac{3}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The normal at (ae, b^2/a) passes through (0, -2b). Using the normal equation (ax/cos theta - by/sin theta = a^2 - b^2) and substituting the point, we solve for e.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If the roots of the equation $x^2 - 4x + 1 = 0$ are the lengths of the semi-major axis and semi-minor axis of an ellipse, then the eccentricity of the ellipse lies between

  1. $\dfrac{1}{3}$ and $\dfrac{1}{2}$
  2. $\dfrac{1}{4}$ and $\dfrac{1}{3}$
  3. $\dfrac{1}{2}$ and $\dfrac{2}{3}$
  4. $\dfrac{2}{3}$ and $1$
Reveal answer Fill a bubble to check yourself
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Explanation

Roots of x^2 - 4x + 1 = 0 are 2 +/- sqrt(3). Thus a = 2 + sqrt(3) and b = 2 - sqrt(3). e^2 = 1 - b^2/a^2. Calculation shows e is between 1/3 and 1/2.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

If $\alpha,\beta$ are the eccentric of the extremities of a focal chord of an ellipse, then eccentricity of the ellipse is

  1. $\dfrac{sin\alpha+sin\beta}{sin(\alpha+\beta)}$
  2. $\dfrac{cos\alpha+cos\beta}{cos(\alpha+\beta)}$
  3. $\dfrac{(\alpha+\beta)}{sin\alpha+sin\beta}$
  4. none of these

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Explanation

The eccentricity of an ellipse given the eccentric angles of the extremities of a focal chord is a standard derivation.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

(-4,1) and (6,1) are the vertices of an ellipse. If one of the foci of the ellipse. If one of the foci of the ellipse lies on x -2y = 2 then its eccentricity is

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

Vertices are (-4,1) and (6,1), so center is (1,1) and 2a = 10, a = 5. Focus lies on x - 2y = 2. With center (1,1), focus is (1+ae, 1). Plugging into x-2y=2: (1+ae) - 2(1) = 2 => ae = 3. e = 3/5.