JEE Parabola | Solved Questions | Coordinate Geometry | Target JEE 2025| JEE Maths

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Till now we have circles and various concepts associated with them, now we will delve into more variety of curves commonly introduced as conic sections. Parabola, the first topic in this chapter of conic sections will be dealt with in this video. Various concepts involved with parabola starting with the definition of the parabola and the terms like focus, vertex, directrix, latus rectum, parametric form of the parabola along with chord and types of chords that can be drawn along with the tangent, normal to the parabola are discussed. We then learn about shifted parabolas whose axis is parallel to the coordinate axis on the contrary to standard parabolas with axis x or y-axis and vertex at the origin.

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Question 1: In this question, we look at the parabola given and make use of the parametric form to define point P and then apply section formula to arrive at the result. 08:45

Question 2: In this question, we look at the circle and the parabola given and use the parametric form of a parabola to represent the intersection and solve both the curves to result in those points. We then define tangents and solve the equations of tangents to arrive at the result. 14:54

Question 3: In this question, we are given a parabola for which a normal drawn cuts parabola again and acts as a chord(hence normal chord). We use the condition of normal chord and equation of the normal to a parabola to arrive at the result. 23:54

Question 4: In this question, we are given a parabola for which two chords are drawn from a given point. In the process of solving the problem, we start with defining these chords by using the parametric form of a parabola and arrive at the result. 31:38

Question 5: In this question, we look at the shifted parabola with axis parallel to the x-axis and a focal chord defined to it. We make use of the condition on the focal chord to arrive at the result. 36:38

Question 6: In this question, we look at the parabola and a line which intersects it. We solve the curves to find the intersection points and then proceed to write the equation of the perpendicular bisector to arrive at the result. 44:28

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