JEECUP (UP Polytechnic) Conic Sections — Ellipse — practice questions
6 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice JEECUP (UP Polytechnic) Conic Sections — Ellipse in the app →For the standard form of a **horizontal** ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $0 < b < a$, 
From the figure, the vertices of the horizontalFind the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.The foci of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ lie at:For an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ ($a > b$), the **length of the latus rectum** is:An ellipse has foci at $(\pm 4, 0)$ and passes through the point $(5, 0)$. Find its standard equation.