Coordinate Geometry Weightage In JEE Mains: Weightage, Marks & Important Topics

Coordinate Geometry Weightage in JEE Main 2027

The total Coordinate Geometry weightage in JEE Main is given below:

Coordinate Geometry has always been one of the highest weightage chapters in the JEE Main syllabus. The chapter has contributed a large number of questions from Straight Lines, Circles, Parabola, Ellipse and Hyperbola in last five years. This is a scoring chapter and has a predictable question pattern so students who are targeting a high percentile in JEE Main 2027 should give this chapter a priority. The weightage of coordinate geometry in the JEE Main 2027 exam is expected to be the highest as it has been in recent years.

Coordinate Geometry Weightage Trend in JEE Main (2021-2026)

Coordinate Geometry Important Questions for JEE Main 2027

Before seeing the coordinate geometry weightage in the JEE Main, let’s first understand the type of questions that are included in this chapter:

Question.1: The area (in sq. units) of the largest rectangle $A B C D$ whose vertices $A$ and $B$ lie on the $x$-axis and vertices C and D lie on the parabola, $y=x^2-1$ below the $x$-axis is:
1) $\frac{2}{3 \sqrt{3}}$
2) $\frac{1}{3 \sqrt{3}}$
3) $\frac{4}{3}$
4) $\frac{4}{3 \sqrt{3}}$

Solution:

$
\begin{aligned}
\text { Area }(A) & =2 t\left(1-t^2\right), \quad(0<t<1) \\
A & =2 t-2 t^3 \\
\frac{d A}{d t} & =2-6 t^2 \\
2-6 t^2 & =0 \\
t & =\frac{1}{\sqrt{3}} \\
\Rightarrow A_{\max } & =2 t\left(1-t^2\right) \\
& =\frac{2}{\sqrt{3}}\left(1-\frac{1}{3}\right) \\
& =\frac{4}{3 \sqrt{3}}
\end{aligned}
$

Question 2: The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2 x-8$, is:
1) 8
2) 9
3) 6
4) 7

Correct Answer: 9

Solution:

$
\begin{aligned}
& y^2=4 x-8 \\
& y=2 x-8 \\
& \Rightarrow y^2=2(y+8)-8 \\
& \Rightarrow y^2-2 y-8=0 \\
& y=4 \quad y=-2 \\
& \int_{-2}^4\left(\frac{y+8}{2}-\frac{y^2+8}{4}\right) d y \\
& \Rightarrow \int_{-2}^4 \frac{y}{2} d y-\int_{-2}^4 \frac{y^2}{4} d y+\int_{-2}^4 2 d y \\
& \Rightarrow\left[\frac{y^2}{4}\right]_{-2}^4-\frac{1}{4}\left[\frac{y^3}{3}\right]_{-2}^4+2[y]_{-2}^4=9
\end{aligned}
$

Question 3: The area enclosed by the curves $x y+4 y=16$ and $x+y=6$ is equal to:
1) $28-30 \log _{\mathrm{e}} 2$
2) $30-28 \log _e 2$
3) $30-32 \log _e 2$
4) $32-30 \log _{\mathrm{e}} 2$

Correct Answer:

$
30-32 \log _e 2
$

Solution:

$
y(x+4)=16 \& y=6-x
$

solve both curves,

$
\begin{aligned}
& (6-x)(x+4)=16 \\
& \int_{-2}^4\left((6-x)-\left(\frac{16}{x+4}\right)\right) d x \\
& \Rightarrow\left[6 x-\frac{x^2}{2}-16 \ln |x+4|\right]_{-2}^4 \\
& \Rightarrow(24-8-16 \ln 8)-(-12-2-16 \ln 2) \\
& \Rightarrow 30-16 \ln 4
\end{aligned}
$

Hence, the answer is option (3).

Question 4: Three points $\mathrm{O}(0,0), \mathrm{P}\left(\mathrm{a}, \mathrm{a}^2\right), \mathrm{Q}\left(-\mathrm{b}, \mathrm{b}^2\right), \mathrm{a}>0, \mathrm{~b}>0$, are on the parabola $y=x^2$. Let $S_1$ be the area of the region bounded by the line $P Q$ and the parabola, and $S_2$ be the area of the triangle OPQ . If the minimum value of $\frac{S_1}{S_2}$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to $\_\_\_\_$ .

Correct Answer: 7
Solution:

$
\begin{aligned}
& y=x^2 \quad P Q ;\left(y-b^2\right)=\frac{a^2-b^2}{a+b}(x+b) \\
& P Q ;(a-b) x-y+a b=0 \\
& \text { Now, } \mathrm{S}_1=\int_{-\mathrm{b}}^{\mathrm{a}}[(\mathrm{a}-\mathrm{b}) \mathrm{x}+\mathrm{ab}]-\left[\mathrm{x}^2\right] \cdot \mathrm{dx} \\
& S_1=\left[(a-b) \frac{x^2}{2}+a b x-\frac{x^3}{3}\right]_{-b}^a \\
& S_1=(a-b) \frac{\left(a^2-b^2\right)}{2}+a b(a+b)-\frac{1}{3}\left[a^3+b^3\right]
\end{aligned}
$

Question 5: The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2 x$ is equal to:
1) $\frac{\pi}{2}-\frac{1}{3}$
2) $\pi-\frac{2}{3}$
3) $\frac{\pi}{2}-\frac{2}{3}$
4) $\pi-\frac{1}{3}$

Correct Answer:

$
\pi-\frac{2}{3}
$

Solution:
Required area $=\operatorname{Ar}($ circle from 0 to 2) $-\operatorname{ar}($ para from 0 to 2)

$
\begin{aligned}
& =\int_0^2 \sqrt{8-x^2} d x-\int_0^2 \sqrt{2 x} d x \\
& =\left[\frac{x}{2} \sqrt{8-x^2}+\frac{8}{2} \sin ^{-1} \frac{x}{2 \sqrt{2}}\right]_0^2-\sqrt{2}\left[\frac{2 x \sqrt{x}}{3}\right]_0^2 \\
& =\frac{2}{2} \sqrt{8-4}+\frac{8}{2} \sin ^{-1} \frac{2}{2 \sqrt{2}}-\frac{2 \sqrt{2}}{3}(2 \sqrt{2}-0) \\
& \Rightarrow 2+4 \cdot \frac{\pi}{4}-\frac{8}{3}=\pi-\frac{2}{3}
\end{aligned}
$

Hence, the answer is option (2)

These are some important questions from coordinate geometry given above. Students can refer to Coordinate Geometry Questions to see more important questions.

Coordinate Geometry Important Topics for JEE Main 2027

The following table has shown the variation of topics asked in coordinate geometry along with which part of the topic you should be focusing upon more.