Embibe Experts Solutions for Chapter: Ellipse, Exercise 1: JEE Advanced Paper 1 - 2020

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Ellipse, Exercise 1: JEE Advanced Paper 1 - 2020

Attempt the free practice questions on Chapter 16: Ellipse, Exercise 1: JEE Advanced Paper 1 - 2020 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Ellipse, Exercise 1: JEE Advanced Paper 1 - 2020 with Hints & Solutions

HARD
JEE Advanced
IMPORTANT

Let E be the ellipse x216+y29=1. For any three distinct points P,Q and Q' on E, let MP,Q be the mid-point of the line segment joining P and Q, and MP,Q' be the mid-point of the line segment joining P and Q'. Then the maximum possible value of the distance between MP,Q and MP,Q', as P,Q and Q' vary on E, is _____.

HARD
JEE Advanced
IMPORTANT

Let a, b and λ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y 2 =4λx , and suppose the ellipse x 2 a 2 + y 2 b 2 =1 passes through the point P . If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is

HARD
JEE Advanced
IMPORTANT

Define the collections E1,E2,E3,.... of ellipses and R1,R2,R3,.... of rectangles as follows:

E1:x29+y24=1;

R1: rectangle of largest area, with sides parallel to the axes, inscribed in E1;

En: ellipse x2an2+y2bn2=1 of largest area inscribed in Rn-1,n>1;

Rn: rectangle of largest area, with sides parallel to the axes, inscribed in En,n>1.

Then which of the following options is/are correct?

HARD
JEE Advanced
IMPORTANT

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1 Column 2 Column 3
(I) x2+y2=a2 (i) my=m2x+a (P) am2,2am
(II) x2+a2y2=a2 (ii) y=mx+a m2+1 (Q) -mam2+1,am2+1
(III) y2=4ax (iii) y=mx+ a2m2-1 (R) -a2ma2m2+1,1a2m2+1
(IV) x2-a2y2=a2 (iv) y=mx+a2m2+1 (S) -a2ma2m2-1,-1a2m2-1
The tangent to a suitable conic (Column 1) at 3,12 is found to be 3x+2y=4 , then which of the following options is the only Correct combination?