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

Coordinate Geometry Weightage In JEE Mains Pdf Free Download

The total 3D Geometry weightage in JEE Mains is given below:

The weightage of coordinate geometry in the JEE Mains 2026 exam is expected to be the highest as it has been in recent years.

Coordinate Geometry last five years Weightage

Also Read:

3D Geometry Weightage in JEE Mains: Important Questions

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

Q. 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:

Option 1:

$$

\frac{2}{3 \sqrt{3}}

$$

Option 2:

$$

\frac{1}{3 \sqrt{3}}

$$

Option 3:

$$

\frac{4}{3}

$$

Option 4:

$$

\frac{4}{3 \sqrt{3}}

$$

Correct Answer:

$$

\frac{4}{3 \sqrt{3}}

$$

Solution:

$$

\begin{aligned}

& \text { Area }(\mathrm{A})=2 \mathrm{t} \cdot\left(1-\mathrm{t}^2\right) \\

& (0<\mathrm{t}<1) \\

& \mathrm{A}=2 \mathrm{t}-2 \mathrm{t}^3 \\

& \frac{\mathrm{dA}}{\mathrm{dt}}=2-6 \mathrm{t}^2 \\

& \mathrm{t}=\frac{1}{\sqrt{3}} \\

& \Rightarrow \mathrm{~A}_{\max }=\frac{2}{\sqrt{3}}\left(1-\frac{1}{3}\right)=\frac{4}{3 \sqrt{3}}

\end{aligned}

$$

Q. 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:

Option 1:

8

Option 2:

9

Option 3:

6

Option 4:

7

Correct Answer:

9

Solution:

$y^2=4 x-8$

$y=2 x-8$

$\Rightarrow y^2=2(y+8)-8$

$$

\begin{aligned}

& \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}

$$

Hence, the answer is the option (2).

Q. 3 The area enclosed by the curves $x y+4 y=16$ and $x+y=6$ is equal to :

Option 1:

$$

28-30 \log _{\mathrm{e}} 2

$$

Option 2:

$$

30-28 \log _e 2

$$

Option 3:

$$

30-32 \log _e 2

$$

Option 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,

$$

(6-x)(x+4)=16

$$

$$

\begin{aligned}

& \int_{-2}^4\left((6-x)-\left(\frac{16}{x+4}\right)\right) d x \\

& \Rightarrow\left[6 x-x^2-16 \ln |x+4|\right]_{-2}^4 \\

& \Rightarrow(24-8-16 \ln 8)-(-12-2-16 \ln 2) \\

& \Rightarrow 30-32 \ln 2

\end{aligned}

$$

Hence, the answer is the option (3).

Q. 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 $\mathrm{y}=\mathrm{x}^2$. Let $\mathrm{S}_1$ be the area of the region bounded by the line PQ 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:

$$

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

$$

Now, $\mathrm{S}_1=\int_{-\mathrm{b}}^{\mathrm{a}}[(\mathrm{a}-\mathrm{b}) \mathrm{x}+\mathrm{ab}]-\left[\mathrm{x}^2\right] \cdot \mathrm{dx}$

$$

\begin{aligned}

& 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}

$$

Solution:

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

$$

\begin{aligned}

& =\int_0^2 \sqrt{8-\mathrm{x}^2} \mathrm{dx}-\int_0^2 \sqrt{2 \mathrm{x}} \mathrm{dx} \\

& =\left[\frac{\mathrm{x}}{2} \sqrt{8-\mathrm{x}^2}+\frac{8}{2} \sin ^{-1} \frac{\mathrm{x}}{2 \sqrt{2}}\right]_0^2-\sqrt{2}\left[\frac{\mathrm{x} \sqrt{\mathrm{x}}}{3 / 2}\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 the option (2)

These are some important questions from coordinate geometry given above. To see more important questions you can use the following link to download.

Co-ordinate Geometry Questions

3D Geometry weightage in JEE Mains: Key Focus Areas and Topics

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