Parabola | Lecture 3 | Practice problems & General parabola | JEE Advanced compendium

Solidify your understanding of parabolas with Lecture 3 of our JEE Advanced Compendium Series! This session is packed with a variety of practice problems and detailed explorations of general parabola concepts. Tailored specifically for JEE aspirants, this video will help you apply theoretical knowledge to solve practical, exam-level questions efficiently. Whether you're looking to reinforce what you've learned or tackle new challenges, this lecture is your essential resource for excelling at parabola questions in the JEE Advanced. Dive in and transform your understanding into top-tier exam performance!
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[@Mathsmerizing]

Errata: At 12:15, the axis should be (x-y)/√2 and (x+y-2)/√2 and that would make A=1/2√2. Since, in the question we had put X=0 and Y=0, the answer to the vertex will not change.

[@immortal_venerable7447]

At 12:50, You directly calculated A of parabola as 1/2 but if we find the focus and the distance between focus and vertex it's not matching, most probably because these lines are parallel to the tangent at vertex and the axis of parabola and not actually them, I transformed the equation into (distance from tangent at vertex)^2=4A(distance from axis of parabola) form and then the value of a matched in both cases. The correct value would be 1/2√2

[@Mathsmerizing]

Yes. The axis should have been (x-y)/√2 and (x+y-2)/√2 and that would make A=1/2√2

[@Koshabhh]

sir blackboard lectures only please. Humble request🙏🙏

[@VS-001]

1:16:38 slight error sir in writing the equating putting value SG. There's an additional a you have put. I tried using angle b/w two lines e to 60. Ans is (d) 4a

[@RekhaDevi-rx4vv]

In the first question, We cannot equate 4A with 14. I plotted it all on a graphing calculator, the directrix corresponding to the parabola is coming out to be wrong as calculated due to the wrong value of A.
I think we should convert it into (dist. From axis)² = LLR(dist. From tangent at vertex) form. From there we are getting correct value of A

[@ShivamTank-pq1jz]

sir 40:48 why only one point of intersection

[@anonymousnill6694]

Sir, at 20:51, how do I imagine the curve y^2=4x being reflected about the line x+y+4=0?

[@kingsqxxd1934]

comman sense

[@taggy9117]

if u can't (like me) then try another solution.
I attempted by finding the mirror parabola's equation.
you can do this by assuming parametric coordinates on parabola and then using image formula.
then just solve with the line y=-5 to get a quadratic equation whose difference of roots will be the answer

[@MRCRAZYSAURABH]

TIMESTAMP 🔥
32:08 *
40:46
43:07

[@immortal_venerable7447]

Thanks

[@MRCRAZYSAURABH]

🔥🔥🔥🔥🔥

[@PasyaUwU]

Yessir