JEE Main Conic Sections — practice questions
31 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice JEE Main Conic Sections in the app →Equation of circle with center (h, k) and radius r:Standard ellipse x²/a² + y²/b² = 1 (a > b) has eccentricity:Hyperbola x²/a² - y²/b² = 1 has eccentricity:Latus rectum of parabola y² = 4ax:Equation of tangent to parabola y² = 4ax at point (at², 2at):Axis of symmetry of parabola x² = 4ay (a > 0):Asymptotes of hyperbola x²/a² - y²/b² = 1:Equation of normal to parabola y² = 4ax at (at², 2at):Conjugate hyperbola of x²/a² - y²/b² = 1:Equation of circle passing through origin, (a, 0), (0, b):Tangent to circle x² + y² = r² at point (x₁, y₁):Find focus and directrix of parabola y² = -8x:Eccentricity of hyperbola 9x² - 16y² = 144:Length of latus rectum of ellipse x²/25 + y²/9 = 1:Common tangent to circle x² + y² = 5 and parabola y² = 8x:Eccentricity of ellipse: x² + 4y² = 4:For parabola y² = 4ax, point (at², 2at) gives parametric form. If t = 2 and a = 3:Sum of distances from any point on ellipse 9x² + 16y² = 144 to its foci:Difference of focal distances from any point on hyperbola x²/16 - y²/9 = 1:For hyperbola x²/25 - y²/144 = 1, find length of conjugate axis:Auxiliary circle of ellipse x²/a² + y²/b² = 1:Relation between eccentricities of conjugate hyperbolas:The equation y² = 4ax represents a PARABOLA with:The eccentricity of a CIRCLE is:The standard form of a circle with centre (3, -2) and radius 5 is:In the standard ellipse x²/a² + y²/b² = 1 (with a > b), the relationship between a, b, c (focal distance) is:The HYPERBOLA x²/a² − y²/b² = 1 has asymptotes given by:For a general 2nd-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0, the curve is a PARABOLA when:By Kepler's First Law, every planet orbits the Sun in:A satellite dish uses a parabolic reflector because:The Cassegrain telescope design uses the reflective property of a: