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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
A Correct answer
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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

For all admissible values of the parameter $a$ the straight line $2ax+y\sqrt{1-a^2}=1$ will touch an ellipse whose eccentricity is equal to

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

The line 2ax + y*sqrt(1-a^2) = 1 is of the form x*cos(theta) + y*sin(theta) = p where cos(theta) = 2a and sin(theta) = sqrt(1-a^2). Squaring and adding gives cos^2(theta) + sin^2(theta) = 4a^2 + 1 - a^2 = 3a^2 + 1 = 1, which is not quite right; however, the condition for a line to touch an ellipse x^2/A^2 + y^2/B^2 = 1 is p^2 = A^2*cos^2(theta) + B^2*sin^2(theta). Comparing coefficients leads to A^2=1/4 and B^2=1, so e^2 = 1 - (1/4)/1 = 3/4, e = sqrt(3)/2.

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

If the focal chord of the ellipse  $\dfrac { x ^ { 2 } } { a ^ { 2 } } + \dfrac { y ^ { 2 } } { b ^ { 2 } } = 1 , ( a > b )$  is normal at  $( a \cos \theta , b \sin \theta )$  then eccentricity of the ellipse is (it is given that  $sin\theta \neq0)$

  1. $| \sec \theta |$
  2. $| \cos \theta |$
  3. $| \sin \theta |$

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

The normal at (a*cos(theta), b*sin(theta)) to the ellipse x^2/a^2 + y^2/b^2 = 1 is ax*sec(theta) - by*csc(theta) = a^2 - b^2. If this is a focal chord, it must pass through a focus (ae, 0). Substituting gives a*ae*sec(theta) = a^2 - b^2. Thus a^2*e*sec(theta) = a^2(1 - b^2/a^2) = a^2*e^2. Simplifying gives e = |sec(theta)|.