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JEE Main 2021 Analysis: Coordinate Geometry Made Easy, Important Questions And Solutions

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By Ramraj Saini

11 Jul'22 4 min read

JEE Main 2021 Analysis: Coordinate Geometry Made Easy, Important Questions And Solutions

Synopsis

Coordinate Geometry is an important part of JEE Main, the tough entrance exam for engineering admission to premier colleges. In JEE Main 2021 some 840 questions were from Mathematics and Coordinate Geometry accounted for 11.07 percent of them. Read through this article for a comprehensive analysis of Coordinate Geometry and know about the most important topics from where maximum questions were asked.

Synopsis

The National Testing Agency (NTA) has been conducting the Joint Entrance Examination (JEE) Main, Paper 1 for admission to undergraduate engineering programs (B.E/B.Tech.) at NITs, IIITs, and other premiere Institutes. Paper 1 is also an eligibility test for JEE Advanced, which is conducted for admission to IITs. Paper 2 is conducted for admission into B.Arch and B.Plan courses across the country.

The JEE Main syllabus consists of three subjects, that is, Physics, Chemistry, and Mathematics, and the Mathematics syllabus contains 16 chapters. JEE Main is a highly competitive exam and so aspirants should have a good command of each topic in every subject to score well in the exam. Mathematics has one-third weightage and therefore it becomes crucial to have a good hold on important topics in Mathematics syllabus. For a dedicated candidate, it is essential to have a winning plan that contains prioritizing important topics and analysis of the previous years’ papers to understand the breadth and depth of the topics.

This article analyzes Coordinate Geometry, one of the important Mathematics chapters of JEE Main 2021. It includes topics like Rectangular Cartesian Coordinate System, Distance Formula, Section Formula, Locus, Straight Lines, Circles, Conic Section, Parabola, Ellipse, Hyperbola, and some other topics as well. This article explains concepts with the help of previous years’ questions.

A total of 28 papers were held in both shifts in all the four sessions of JEE Main 2021. In these papers, 840 questions were from mathematics. Out of them 93 questions, that is, 11.07%, were from Coordinate Geometry. It shows how important this chapter is for JEE Main aspirants. The concepts, number of questions asked and weightage is given in the following table.

Concepts And Number Of Questions

Topic	Number Of Questions	Weightage (%)
Circle	35	37.63
Parabola	17	18.27
Ellipse	12	12.90
Hyperbola	8	8.60
Line	42	45.16
Tangent	32	34.40
Normal	9	9.67
Locus	14	15.05
Area Of Triangle	11	11.82

It can be observed that out of 93 questions, concepts of a Line, Circle, Tangent are used in 42, 35 and 32 questions respectively. Concepts of the Circle contain standard and general forms of the equation of a circle, the equation of a circle when the endpoint of diameter is given, the intersection of circle and line, etc. Concepts of the line include an intersection of lines, the angle between two lines, condition for the intersection of three lines, coordinate of centroid, circumcenter, the orthocenter of a triangle, family of lines.

Solution of the following few questions from JEE Main 2021 show that command over multiple concepts is required to solve a problem. Practising every concept is essential as it can be difficult for students who are not aware or do not have a good hold on all the concepts to solve such questions.

Q.1 JEE Main 2021

The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d² is equal to

Solution

Let the point be P (h,k)

As per the question

$h^{2}+k^{2}+\left ( h-1 \right )^{2}+k^{2}+h^{2}+\left ( k-1 \right )^{2}+\left ( h-1 \right )^{2}+\left ( k-1 \right )^{2}= 18$

$\Rightarrow 4h^{2}+4k^{2}-4h-4k= 14$

$\Rightarrow x^{2}+y^{2}-x-y-\frac{7}{2}= 0$

$Diameter\, d= 2\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{7}{2}}$

$= 2\sqrt{\frac{8}{2}}$

Hence,

$d^{2}= 16$

Concepts Used

Distance formula, Concepts of locus, Equation of a circle.

Q.2 JEE Main 2021

Consider the parabola with vertex $\left(\frac{1}{2}, \frac{3}{4}\right)$ and the directrix $y=\frac{1}{2}$ . Let be the point where the parabola meets the line $x=-\frac{1}{2}$ . If the normal to the parabola at $\mathrm{P}$ intersects the parabola again at the point $\mathrm{Q}$ , then $(\mathrm{PQ})^{2}$ is equal to.

Solution

$clearly\: a= \frac{1}{4}$

Equation of parabola

$\left ( x-\frac{1}{2} \right )^{2}= 4\left ( \frac{1}{4} \right )\left ( y-\frac{3}{4} \right )$

For point P

$\left ( -\frac{1}{2}-\frac{1}{2} \right )^{2}= \left ( q-\frac{3}{4} \right )\Rightarrow q= \frac{7}{4}$

For PQ

$2\left ( x-\frac{1}{2} \right )= {y}'$

$At\: P,{y}'= -2$

$\therefore slope\: of\: normal= \frac{1}{2}$

Equation of PQ:

$y-\frac{7}{4}= \frac{1}{2}\left ( x+\frac{1}{2} \right )$

$\Rightarrow y= \frac{x}{2}+\frac{1}{4}+\frac{7}{4}= \frac{x}{2}+2$

For point Q

$\left ( x-\frac{1}{2} \right )^{2}= \left ( \frac{x}{2} +2-\frac{3}{4}\right )$

$\Rightarrow x= 2$

$\therefore Q= \left ( 2,3 \right )$

$PQ^{2}= \left ( 2+\frac{1}{2} \right )^{2}+\left ( 3-\frac{7}{4} \right )^{2}= \frac{125}{16}$

Concepts Used

Intersection of line and parabola, Equation of a parabola, Concepts of the normal, Equation of a line in slope-intercept form, Distance formula.

Q.3 JEE Main 2021

The locus of mid-points of the line segments joining and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ are:

$36 x^{2}+16 y^{2}+90 x+56 y+145=0$
$36 x^{2}+16 y^{2}+108 x+80 y+145=0$
$9 x^{2}+4 y^{2}+18 x+8 y+145=0$
$36 x^{2}+16 y^{2}+72 x+32 y+145=0$

Solution

Let us take any parametric point $B\left ( 2\cos\theta,3\sin\theta \right )$ on the ellipse

Let be the midpoint of

So $h=\frac{2\cos\theta-3}{2}\Rightarrow \cos\theta= \frac{2h+3}{2}\\$

$k=\frac{3\sin\theta-5}{2}\Rightarrow \sin\theta=\frac{2k+5}{3}\\$

Square and add;

$\cos^{2}\theta+\sin^{2}\theta=\left ( \frac{2h+3}{2} \right )^{2}+\left ( \frac{2k+5}{3} \right )^{2}=1\\$

Replace $h\rightarrow x\&\; k\rightarrow y\\$

$\Rightarrow \frac{4x^{2}+12x+9}{4}+\frac{4y^{2}+20y+25}{9}=1\\$

$\Rightarrow 36x^{2}+108x+81+16y^{2}+80y+100-36\\$

$\Rightarrow 36x^{2}+16y^{2}+108x+80y+145=0$

Concepts Used

Point on the ellipse in parametric form, Concepts of locus.

Q.4 JEE Main 2021

The locus of the midpoints of the chords of the hyperbola $x^{2}-y^{2}= 4,$ which touch the parabola $y^{2}= 8x,$ is:

$y^{2}\left ( x-2 \right )= x^{3}$
$x^{3}\left ( x-2 \right )= y^{2}$
$x^{2}\left ( x-2 \right )= y^{3}$
$y^{3}\left ( x-2 \right )= x^{2}$

Solution

Let the midpoint of chord be P(h,k)

$\therefore$ equation of chord: T = S1

$hx-ky-4 = h^{2}-k^{2}-4$

$\Rightarrow hx-ky = h^{2}-k^{2}$

$\Rightarrow ky = hx-h^{2}+k^{2}$

$\Rightarrow y =\left ( \frac{h}{k} \right )x+\left ( \frac{k^{2}-h^{2}}{k} \right )----(i)$

Any tangent to $y^{2}= 8x\: \: \: is\,$

$y= mx+\frac{2}{m}$

Comparing it with (i)

$m=\frac{h}{k}\: \: and\: \: \frac{k^{2}-h^{2}}{k}= \frac{2}{m}$

$\Rightarrow \frac{k^{2}-h^{2}}{k}= \frac{2k}{h}$

$\Rightarrow h\left ( k^{2}-h^{2} \right )= 2k^{2}$

$\Rightarrow hk^{2}-h^{3}= 2k^{2}$

$\Rightarrow k^{2}\left ( h-2 \right )= h^{3}$

$\Rightarrow y^{2}\left ( x-2 \right )= x^{3}$

Concepts Used

Equation of cord, Tangent on a parabola, Concepts of locus

Q.5 JEE Main 2021

The point $P\left ( -2\sqrt{6},\sqrt{3} \right )$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1$ having eccentricity $\frac{\sqrt{5}}{2}.$ If the tangent and normal at P to the hyperbola intersects its conjugate axis at the points Q and R respectively, then QR is equal to

$4\sqrt{3}$
$3\sqrt{6}$
$6\sqrt{3}$

Solution

$e^{2}= 1+\frac{b^{2}}{a^{2}}\Rightarrow \frac{5}{4}= 1+\frac{b^{2}}{a^{2}}\Rightarrow \frac{b^{2}}{a^{2}}= \frac{1}{4}$

and $\left ( -2\sqrt{6},\sqrt{3} \right )$ lies on it, so

$\frac{24}{a^{2}}-\frac{3}{b^{2}}= 1\Rightarrow \frac{24}{a^{2}}-\frac{3\cdot 4}{a^{2}}= 1$

$\Rightarrow \frac{1}{a^{2}}\left ( 12 \right )= 1\Rightarrow a^{2}= 12$

$\Rightarrow b^{2}= 3$

Equation of the tangent at P:

T = 0

$\frac{-2\sqrt{6}\cdot x}{12}-\frac{\sqrt{3}y}{3}= 1\Rightarrow slope= \frac{-1}{\sqrt{2}}$

$For\, Q\, put\: x= 0 \Rightarrow y= -\sqrt{3}$

$Q\left ( 0,-\sqrt{3} \right )$

Normal at p

$y-\sqrt{3}= \sqrt{2}\left ( x+2\sqrt{6} \right )$

$For\, R, \, put\, x= 0\Rightarrow y= 5\sqrt{3}.$

$R\left ( 0,5\sqrt{3} \right )$

$\therefore QR= 6\sqrt{3}$

Concepts Used

Equation of hyperbola, a point on hyperbola, tangent and normal on hyperbola, distance formula.

An analysis of the previous year's question shows that Coordinate Geometry questions can be solved very easily by having command over the basic concepts of each topic. Practising as many questions as possible, doing calculations fast without making any errors and practising NCERT book problems are also important. Lastly, solving problems from previous years’ JEE Main exam is highly beneficial to ace the JEE Main exam as it will let you know beforehand what sort of questions to expect and that can give you an edge over other competitors.

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Exam Prep

JEE Main 2021 Analysis: Coordinate Geometry Made Easy, Important Questions And Solutions

By Ramraj Saini

11 Jul'22 4 min read

Synopsis

Synopsis

Concepts And Number Of Questions

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