Mathematics

Conic Sections

285 Questions

Conic sections deal with the geometry of curves like ellipses, hyperbolas, and parabolas formed by the intersection of a plane with a cone. This is an important topic in advanced mathematics sections of competitive exams. Practice these questions to understand eccentricity, focal distances, and equations of tangents and normals.

Ellipse equationsHyperbola eccentricityFocal distance calculationsTangent and normal equationsConic standard forms

Conic Sections Questions

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

The ellipse $E _1:\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ is inscribed in a rectangle R whose sides are parallel to the coordinates axis. Another ellipse $E _2$ passing through the point $(0, 4)$ circumscribes the rectangle R. The eccentricity of the ellipse $E _2$ is?

  1. $\dfrac{\sqrt{2}}{2}$
  2. $\dfrac{\sqrt{3}}{2}$
  3. $\dfrac{1}{2}$
  4. $\dfrac{3}{4}$
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Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

Eccentricity of the ellipse $5x^{2}+6xy+5y^{2}=8$ is

  1. $\dfrac {1}{\sqrt {2}}$
  2. $\dfrac {\sqrt {3}}{2}$
  3. $\sqrt {\dfrac {2}{3}}$
  4. $\dfrac {1}{\sqrt {3}}$
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Explanation

For 5x^2 + 6xy + 5y^2 = 8, rotate the axes to eliminate the xy term. The eigenvalues of the matrix determine the semi-axes, leading to e = 1/sqrt(2).

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

An ellipse has $OB$ as its semi-minor axis. $F _{1}$ and $F _{2}$ are its foci and angle $F _{1}BF _{2}$ is a right angle. The eccentricity of the ellipse is 

  1. $1/\sqrt{2}$
  2. $1/2$
  3. $1/\sqrt{3}$
  4. $2/\sqrt{3}$
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Explanation

If angle F1BF2 = 90 degrees, then in the right triangle F1OB, OB = OF1 = ae. Since b = ae, and b^2 = a^2(1-e^2), we get a^2e^2 = a^2(1-e^2), so 2e^2 = 1.

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

The tangent at any point $P\left(a\cos\theta,b\sin\theta\right)$ on the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ meets the auxiliary circle at two points which subtend a right angle at the center ,then eccentricity is 

  1. $\dfrac{1}{\sqrt{1+\sin^{2}\theta}}$
  2. $\dfrac{1}{\sqrt{2-\cos^{2}\theta}}$
  3. $\dfrac{1}{\sqrt{1+\tan^{2}\theta}}$
  4. $none\ of\ these$
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Explanation

The condition that the tangent meets the auxiliary circle at points subtending 90 degrees at the center relates the coordinates to the eccentricity.

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 } } $
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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. $\dfrac{3}{4}$
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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 $1$
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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. $\dfrac{1}{5}$
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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.