Original JEE Main-style questions on straight lines and conic sections, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 11, Straight Lines · Class 11, Conic Sections.
The distance between the parallel lines \(5x - 12y + 4 = 0\) and \(10x - 24y - 18 = 0\) is
Write the second line as \(5x - 12y - 9 = 0\) (same \(x, y\) coefficients as the first).
Distance \(= \dfrac{|4 - (-9)|}{\sqrt{25 + 144}} = \dfrac{13}{13} = 1\).
Trap: Using \(|c_1 - c_2|\) before making the \(x\) and \(y\) coefficients equal gives \(22/13\).
The eccentricity of the ellipse \(\dfrac{x^2}{169} + \dfrac{y^2}{144} = 1\) is
\(a^2 = 169\), \(b^2 = 144\), and \(b^2 = a^2(1 - e^2)\): \(1 - e^2 = \dfrac{144}{169}\), so \(e^2 = \dfrac{25}{169}\) and \(e = \dfrac{5}{13}\).
Trap: \(25/169\) is \(e^2\): take the square root.
Find the length of the chord that the \(x\)-axis cuts off on the circle \(x^2 + y^2 - 4x + 6y - 12 = 0\).
Put \(y = 0\): \(x^2 - 4x - 12 = 0\), so \((x - 6)(x + 2) = 0\) and \(x = 6\) or \(-2\).
Chord length \(= 6 - (-2) = 8\).
CBSE Essentials opens them with worked solutions, along with every chapter question and marking scheme for Class 10, 11 and 12. Try it free for 7 days, no card.
Start free 7-day trial Sign inAll questions are original, written by CBSE Math Revision in the style of each exam; none are taken from NTA, IIT, CBSE or NCERT papers. Every answer was checked twice: by a computer re-solve and by hand. No calculator needed. Spotted a mistake? Tell us.