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OJEE Conic Sections — practice questions

Explore 89 free Conic Sections questions in OJEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Conic Sections questions and exercises with answers

Showing questions 1–30 of 89.

Find the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.Read the answer and worked solution →The foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ lie at:Read the answer and worked solution →An ellipse has foci at $(\pm 4, 0)$ and passes through the point $(5, 0)$. Find its standard equation.Read the answer and worked solution →Find the length of the latus rectum of the parabola $y^2 = 8x$.Read the answer and worked solution →Find the equation of the parabola with vertex at the origin and focus at $(0, -3)$.Read the answer and worked solution →A parabola has its vertex at the origin, its axis along the positive $x$-axis, and passes through the point $(4, 6)$. Find its equation.Read the answer and worked solution →For the parabola $y^2 = 16x$, find the equation of the directrix.Read the answer and worked solution →The standard equation of a hyperbola with centre at the origin and foci on the $x$-axis is:Read the answer and worked solution →For any hyperbola, the eccentricity $e$ satisfies:Read the answer and worked solution →What is the eccentricity of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$?Read the answer and worked solution →A hyperbola has foci at $(\pm 5, 0)$ and the length of the transverse axis is $6$. Find its equation.Read the answer and worked solution →$\sin(180° - \theta)$ equals:Read the answer and worked solution →$\sin^2\theta + \cos^2\theta$ equals:Read the answer and worked solution →$\sin(A + B) = $Read the answer and worked solution →sin(0°) equals:Read the answer and worked solution →cos(90°) equals:Read the answer and worked solution →sin²θ + cos²θ equals:Read the answer and worked solution →tan(45°) equals:Read the answer and worked solution →In which quadrant is sin θ positive and cos θ negative?Read the answer and worked solution →Equation of circle with center (h, k) and radius r:Read the answer and worked solution →Standard ellipse x²/a² + y²/b² = 1 (a > b) has eccentricity:Read the answer and worked solution →Hyperbola x²/a² - y²/b² = 1 has eccentricity:Read the answer and worked solution →Latus rectum of parabola y² = 4ax:Read the answer and worked solution →Equation of tangent to parabola y² = 4ax at point (at², 2at):Read the answer and worked solution →Axis of symmetry of parabola x² = 4ay (a > 0):Read the answer and worked solution →Asymptotes of hyperbola x²/a² - y²/b² = 1:Read the answer and worked solution →Equation of normal to parabola y² = 4ax at (at², 2at):Read the answer and worked solution →Conjugate hyperbola of x²/a² - y²/b² = 1:Read the answer and worked solution →Equation of circle passing through origin, (a, 0), (0, b):Read the answer and worked solution →Tangent to circle x² + y² = r² at point (x₁, y₁):Read the answer and worked solution →

Worked practice questions

Find the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.

  1. $\dfrac{3}{4}$
  2. $\dfrac{3}{5}$
  3. $\dfrac{4}{5}$
  4. $\dfrac{5}{4}$
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Correct answer: B. $\dfrac{3}{5}$

Identify: $a^2 = 25$ (the larger denominator), $b^2 = 16$. So $a = 5, b = 4$. $c^2 = a^2 - b^2 = 25 - 16 = 9 \Rightarrow c = 3$. Eccentricity $e = \dfrac{c}{a} = \dfrac{3}{5}$. For any ellipse $0 < e < 1$ (a circle is the $e = 0$ limit). Option D ($5/4 > 1$) describes a *hyperbola*, not an ellipse.

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The foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ lie at:

  1. $(\pm 4, 0)$
  2. $(\pm 3, 0)$
  3. $(0, \pm 3)$
  4. $(\pm 5, 0)$
Show answer and solution

Correct answer: B. $(\pm 3, 0)$

For a horizontal ellipse the foci sit at $(\pm c, 0)$ on the major axis, where $c^2 = a^2 - b^2$. $c^2 = 25 - 16 = 9 \Rightarrow c = 3$. So the foci are at $(\pm 3, 0)$. Option D mistakenly picks the *vertices* option A switches $a$ and $b$.

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An ellipse has foci at $(\pm 4, 0)$ and passes through the point $(5, 0)$. Find its standard equation.

  1. $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$
  2. $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$
  3. $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$
  4. $\dfrac{x^2}{9} + \dfrac{y^2}{16} = 1$
Show answer and solution

Correct answer: A. $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$

**Read the geometry.** Foci on the $x$-axis at $(\pm 4, 0)$ ⇒ horizontal ellipse with $c = 4$. The point $(5, 0)$ lies on the major axis ($y = 0$) and on the curve ⇒ it must be a *vertex*, giving $a = 5$. **Recover $b$.** $b^2 = a^2 - c^2 = 25 - 16 = 9$, so $b = 3$. **Equation:** $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$. **Distractor check.** Option B has $b = 4$ — that would put the foci at $(\pm 3, 0)$, not the given $(\pm 4, 0)$.

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