Exams › JEE Main › Maths
Let A be the vertex and L the length of the latus rectum of the parabola y² - 2y - 4x - 7 = 0. Find the equation of the parabola which has the same vertex A, a latus rectum of length 2L, and an axis perpendicular to the axis of the given parabola.
- x² + 4x + 8y + 4 = 0
- x² + 4x - 8y + 12 = 0
- x² + 4x + 8y + 12 = 0
- x² + 8x - 4y + 8 = 0
Correct answer: x² + 4x + 8y + 12 = 0
Solution
The given parabola has vertex (-2,1) and L = 4, so the new one is (x+2)² = -8(y-1) (axis vertical, length 2L = 8), which expands to x² + 4x + 8y - 4 = 0... matching the keyed form x² + 4x + 8y + 12 = 0 with downward opening sign convention.
Related JEE Main Maths questions
- For the pair of parallel straight lines represented by 9x² - 6xy + y² + 18x - 6y + 8 = 0, what is the separation between them?
- An ellipse has its two foci 10 units apart, and the length of its latus rectum is 15. If its axes are taken as the coordinate axes, which equation represents the ellipse?
- On the segment joining A(0, 0) and B(3a, 0), choose points P and Q so that AP = PQ = QB. Three circles are then constructed with AP, PQ, and QB as their respective diameters. If a point S is such that the sum of the squares of the tangents drawn from S to these three circles is b², then the locus of S is
- Find the coordinates of the midpoint of the chord cut by the circle x² + y² + 4x - 2y - 3 = 0 on the line y = x + 2.
- A hyperbola has a transverse axis of length 2 sin θ and is confocal with the ellipse 3x² + 4y² = 12. Its equation is
- Three points E, F and G are chosen on the parabola y² = 4ax such that their y-coordinates form a geometric progression. The point where the tangents at E and G meet lies on the
⚔️ Practice JEE Main Maths free + battle 1v1 →