How much of JEE Main comes from Coordinate Geometry?

JEEGenie estimates about 8% of JEE Main marks come from Coordinate Geometry, based on recent papers. The share varies from year to year.

Which formulas matter most in Coordinate Geometry?

The Coordinate Geometry sheet on this page has 23 formula entries, starting with Coordinate System & Distance Formula, Section Formula, Area of a Triangle, Slope of a Line, Equation of a Line: Various Forms. Each one comes with when to use it and the trap to avoid.

What topics does Coordinate Geometry cover for JEE?

Coordinate Geometry covers Straight Lines, Circles, Parabola, Ellipse, Hyperbola, Pair of Straight Lines.

How should I revise Coordinate Geometry before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Coordinate Geometry questions on JEEGenie.

How many previous-year questions are there on Coordinate Geometry?

JEEGenie has 1167 previous-year questions on Coordinate Geometry, and this page shows a few with worked explanations.

Is practice on Coordinate Geometry free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Coordinate Geometry for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 8% of JEE Main marks come from Coordinate Geometry. This page covers 23 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Forgetting to square the differences before adding.

Key points to know

Coordinate System & Distance Formula. The distance between two points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ is $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ .
Section Formula. For point $P$ dividing $A B$ in ratio $m : n$ , coordinates are $(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$ for internal division; for external, use $m : - n$ .
Area of a Triangle. Area of triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is $\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ .
Slope of a Line. Slope $m$ of line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , representing steepness and direction.
Locus. Locus is the set of all points satisfying a given geometric condition, expressed as an equation in $x$ and $y$ .
Circle: Standard Equation. Circle with center $(h, k)$ and radius $r$ has equation $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
Parabola: Standard Equation. Standard parabola $y^{2} = 4 a x$ opens right, focus at $(a, 0)$ , directrix $x = - a$ , vertex at origin.
Ellipse: Standard Equation. Standard ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $a > b$ has foci at $(\pm a e, 0)$ , eccentricity $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$ .
Hyperbola: Standard Equation. Standard hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ has foci at $(\pm a e, 0)$ , eccentricity $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$ , asymptotes $y = \pm \frac{b}{a} x$ .
Rectangular Hyperbola. Rectangular hyperbola has $a = b$ , so $x^{2} - y^{2} = a^{2}$ or $x y = c^{2}$ ; asymptotes are perpendicular.

Formula sheet

Name	Formula	When to use
Coordinate System & Distance Formula	$P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$	Use to find the length of a line segment between two points, or to check collinearity (sum of two distances equals third).
Section Formula	Internal: $P = (\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$ ; External: $P = (\frac{m x_{2} - n x_{1}}{m - n}, \frac{m y_{2} - n y_{1}}{m - n})$	Use to find coordinates of a point dividing a segment in a given ratio, or to find centroid, incenter, or harmonic conjugates.
Area of a Triangle	$Δ = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$	Use to compute area from coordinates, or to check collinearity (area = 0).
Slope of a Line	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \tan θ$	Use to find line equations, check parallelism/perpendicularity, or find angle between lines.
Equation of a Line: Various Forms	$y = m x + c$ ; $y - y_{1} = m (x - x_{1})$ ; $\frac{x}{a} + \frac{y}{b} = 1$ ; $A x + B y + C = 0$	Use appropriate form based on given data: slope and intercept, point and slope, two points, or intercepts.
Angle Between Two Lines	$\tan θ = ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣$	Use to find acute angle between two lines, or to check if lines are parallel or perpendicular.
Distance from a Point to a Line	$d = \frac{∣ A x_{1} + B y_{1} + C ∣}{\sqrt{A^{2} + B^{2}}}$	Use to find shortest distance from a point to a line, or in locus problems.
Distance Between Two Parallel Lines	$d = \frac{∣ C_{1} - C_{2} ∣}{\sqrt{A^{2} + B^{2}}}$	Use to find distance between two parallel lines, e.g., in problems involving strips or loci.
Locus	No single formula; derived from condition.	Use to find equation of a curve defined by a geometric property (e.g., equidistant from two points).
Circle: Standard Equation	$(x - h)^{2} + (y - k)^{2} = r^{2}$	Use to represent a circle given center and radius, or to find center and radius from general form.
Equation of Circle: Diameter Form	$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$	Use when given endpoints of a diameter to quickly write circle equation.
Tangent to a Circle	$T = 0$ : $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$	Use to find equation of tangent at a given point on the circle.
Length of Tangent from External Point	$L = \sqrt{S_{1}}$ where $S_{1} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$	Use to find length of tangent from an external point to a circle.
Parabola: Standard Equation	$y^{2} = 4 a x$ (right); $y^{2} = - 4 a x$ (left); $x^{2} = 4 a y$ (up); $x^{2} = - 4 a y$ (down)	Use to identify parabola orientation and key parameters from equation.
Tangent to Parabola $y^{2} = 4 a x$	$y y_{1} = 2 a (x + x_{1})$ ; $y = m x + \frac{a}{m}$	Use to find equation of tangent to parabola at a point or with given slope.
Ellipse: Standard Equation	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ; $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$	Use to represent ellipse and find foci, vertices, eccentricity.
Tangent to Ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$	$\frac{x x_{1}}{a^{2}} + \frac{y y_{1}}{b^{2}} = 1$ ; $y = m x \pm \sqrt{a^{2} m^{2} + b^{2}}$	Use to find tangent equation to ellipse at a point or with given slope.
Hyperbola: Standard Equation	$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ ; $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$ ; asymptotes: $y = \pm \frac{b}{a} x$	Use to represent hyperbola and find foci, vertices, asymptotes.
Tangent to Hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$	$\frac{x x_{1}}{a^{2}} - \frac{y y_{1}}{b^{2}} = 1$ ; $y = m x \pm \sqrt{a^{2} m^{2} - b^{2}}$	Use to find tangent equation to hyperbola at a point or with given slope.
Rectangular Hyperbola	$x y = c^{2}$ (rotated form); $x^{2} - y^{2} = a^{2}$	Use when hyperbola has equal axes or equation $x y = c^{2}$ .
Condition for Tangency: Line to Conic	For parabola $y^{2} = 4 a x$ : $c = a / m$ ; for ellipse: $c^{2} = a^{2} m^{2} + b^{2}$ ; for hyperbola: $c^{2} = a^{2} m^{2} - b^{2}$	Use to find condition for a line to be tangent to a conic, or to find tangent lines with given slope.
Pole and Polar	For circle: $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$	Use to find polar of a point with respect to a conic, or to find pole of a line.
Chord of Contact	For circle: $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$	Use to find equation of chord joining points of tangency from an external point.

Open the Coordinate Geometry formula page

What JEE asks

Coordinate System & Distance Formula. JEE often asks to compute distance in coordinate geometry problems, or use it to derive equations of circles or loci.
Section Formula. JEE tests this in problems involving medians, angle bisectors, or finding points on lines.
Area of a Triangle. JEE often uses this to find area of polygons by splitting into triangles, or in locus problems.
Slope of a Line. JEE tests slope in conjunction with line equations and conic sections.
Equation of a Line: Various Forms. JEE often requires converting between forms or using the general form for distance and angle calculations.
Angle Between Two Lines. JEE tests this in problems involving angle bisectors or conic tangents.
Distance from a Point to a Line. JEE often uses this in problems involving circles (distance from center to tangent) or conics.
Distance Between Two Parallel Lines. JEE tests this in coordinate geometry problems with parallel lines.
Locus. JEE often asks to find locus of a point moving under a condition, e.g., sum of distances constant (ellipse).
Circle: Standard Equation. JEE tests circle equations in problems involving tangents, chords, and intersections with lines.
Equation of Circle: Diameter Form. JEE often uses this in problems involving right angles or chords.
Tangent to a Circle. JEE tests this in problems involving tangents from external points or length of tangent.
Length of Tangent from External Point. JEE often asks to find length of tangent or use it in locus problems.
Parabola: Standard Equation. JEE tests parabola in problems involving tangents, normals, and chord properties.
Tangent to Parabola $y^{2} = 4 a x$ . JEE often asks to find tangent equations or use them in intersection problems.
Ellipse: Standard Equation. JEE tests ellipse in problems involving tangents, normals, and area.
Tangent to Ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ . JEE often asks to find tangent equations or use them in intersection with lines.
Hyperbola: Standard Equation. JEE tests hyperbola in problems involving tangents, normals, and rectangular hyperbola.
Tangent to Hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . JEE often asks to find tangent equations or use them in intersection problems.
Rectangular Hyperbola. JEE tests rectangular hyperbola in problems involving tangents and normals.
Condition for Tangency: Line to Conic. JEE often asks to find tangents from external points or common tangents.
Pole and Polar. JEE tests pole-polar in problems involving chords of contact and harmonic conjugates.
Chord of Contact. JEE often asks to find chord of contact or its length.

Common traps

Forgetting to square the differences before adding. Coordinate System & Distance Formula
Mixing up $m$ and $n$ with the points; for $A P : P B = m : n$ , $m$ multiplies $B$ and $n$ multiplies $A$ (each point is weighted by the opposite segment). Section Formula
Forgetting the absolute value; area is always positive. Area of a Triangle
Dividing by zero when $x_{1} = x_{2}$ (vertical line, slope undefined). Slope of a Line
Using slope-intercept for vertical lines (slope undefined). Equation of a Line: Various Forms
Forgetting the absolute value; angle is acute unless specified. Angle Between Two Lines
Forgetting the absolute value in numerator. Distance from a Point to a Line
Forgetting that lines must be in the same form (coefficients of $x$ and $y$ proportional). Distance Between Two Parallel Lines
Forgetting to eliminate parameters when multiple conditions exist. Locus
Forgetting to square the radius. Circle: Standard Equation
Forgetting that this is derived from the condition that angle in a semicircle is 90°. Equation of Circle: Diameter Form
Forgetting to replace $x^{2}$ with $x x_{1}$ , $y^{2}$ with $y y_{1}$ , $x$ with $\frac{x + x_{1}}{2}$ , etc. Tangent to a Circle
Forgetting that $S_{1}$ must be positive for external point. Length of Tangent from External Point
Confusing focus and directrix: for $y^{2} = 4 a x$ , focus is $(a, 0)$ , not $(0, a)$ . Parabola: Standard Equation

Topics in Coordinate Geometry

Straight Lines
Circles
Parabola
Ellipse
Hyperbola
Pair of Straight Lines

Previous-year questions on Coordinate Geometry

JEEGenie has 1167 previous-year questions on Coordinate Geometry. A few worked examples follow.

Year	Questions
2026	71
2025	74
2024	79
2023	74
2022	97
2021	101
2020	51
2019	110
2018	32
2017	21
2016	32
2015	25
2014	26
2013	34
2012	29
2011	12
2010	15
2009	12
2008	12
2007	18
2006	14
2005	15
2004	18
2003	15
2002	16
2001	10
2000	9
1999	8
1998	13
1997	8
1996	9
1995	6
1994	4
1993	6
1992	6
1991	4
1990	4
1989	5
1988	8
1987	2
1986	8
1985	8
1984	8
1983	12
1982	4
1981	4
1980	8
1979	3
1978	7

Question 1 medium

Let the mid points of the sides of a triangle ABC be

(\frac{5}{2}, 7)

(\frac{5}{2}, 3)

and

(4, 5)

. If its incentre is

(h, k)

, then

3 h + k

is equal to :

Answer: C

Explanation. Let the midpoints be

D (\frac{5}{2}, 7), E (\frac{5}{2}, 3), F (4, 5)

where

D

is midpoint of

B C

E

C A

F

A B

. Using

A = D + E - F, B = D + F - E, C = E + F - D

we get:

A = (\frac{5}{2} + \frac{5}{2} - 4, 7 + 3 - 5) = (1, 5)

B = (\frac{5}{2} + 4 - \frac{5}{2}, 7 + 5 - 3) = (4, 9)

C = (\frac{5}{2} + 4 - \frac{5}{2}, 3 + 5 - 7) = (4, 1)

Now side lengths:

a = B C = ∣ 9 - 1 ∣ = 8

b = C A = \sqrt{(4 - 1)^{2} + (1 - 5)^{2}} = \sqrt{9 + 16} = 5

c = A B = \sqrt{(4 - 1)^{2} + (9 - 5)^{2}} = \sqrt{9 + 16} = 5

Incenter coordinates:

h = \frac{a x_{A} + b x_{B} + c x_{C}}{a + b + c} = \frac{8 (1) + 5 (4) + 5 (4)}{8 + 5 + 5} = \frac{8 + 20 + 20}{18} = \frac{48}{18} = \frac{8}{3}

k = \frac{a y_{A} + b y_{B} + c y_{C}}{a + b + c} = \frac{8 (5) + 5 (9) + 5 (1)}{18} = \frac{40 + 45 + 5}{18} = \frac{90}{18} = 5

Thus:

3 h + k = 3 (\frac{8}{3}) + 5 = 8 + 5 = 13

C

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 hard

A line passing through the point

P (\sqrt{5}, \sqrt{5})

intersects the ellipse

\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1

A

and

B

such that

(P A) \cdot (P B)

is maximum. Then

5 (P A^{2} + P B^{2})

is equal to :

Answer: C

Explanation. Let the line through

P (\sqrt{5}, \sqrt{5})

have direction cosines

(\cos θ, \sin θ)

. A point on the line at signed distance

t

from

P

(x, y) = (\sqrt{5} + t \cos θ, \sqrt{5} + t \sin θ) .

Substitute into the ellipse:

\frac{(\sqrt{5} + t \cos θ)^{2}}{36} + \frac{(\sqrt{5} + t \sin θ)^{2}}{25} = 1.

Expanding,

(\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25}) t^{2} + 2 \sqrt{5} (\frac{\cos θ}{36} + \frac{\sin θ}{25}) t + (\frac{5}{36} + \frac{5}{25} - 1) = 0.

Now

\frac{5}{36} + \frac{5}{25} - 1 = \frac{5}{36} + \frac{1}{5} - 1 = \frac{25 + 36 - 180}{180} = - \frac{119}{180} .

So the quadratic in

t

(\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25}) t^{2} + 2 \sqrt{5} (\frac{\cos θ}{36} + \frac{\sin θ}{25}) t - \frac{119}{180} = 0.

Its roots are the signed distances

P A

and

P B

, say

t_{1}, t_{2}

. Hence

P A \cdot P B = ∣ t_{1} t_{2} ∣ .

By Vieta,

t_{1} t_{2} = \frac{- \frac{119}{180}}{\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25}} = - \frac{119}{180 (\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25})} .

Thus

P A \cdot P B = \frac{119}{180 (\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25})} .

To maximize

P A \cdot P B

, minimize

\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25} .

Write

\frac{\cos^{2} θ}{36} + \frac{\sin^{2} θ}{25} = \frac{1 - \sin^{2} θ}{36} + \frac{\sin^{2} θ}{25} = \frac{1}{36} + \sin^{2} θ (\frac{1}{25} - \frac{1}{36}) .

Since

\frac{1}{25} - \frac{1}{36} = \frac{11}{900} > 0,

this expression is minimized when

\sin^{2} θ = 0

, i.e.

\cos^{2} θ = 1

. Therefore the maximizing line is horizontal. For

\cos^{2} θ = 1, \sin^{2} θ = 0

, the quadratic becomes

\frac{1}{36} t^{2} + 2 \sqrt{5} \cdot \frac{1}{36} t - \frac{119}{180} = 0.

Multiply by

180

5 t^{2} + 10 \sqrt{5} t - 119 = 0.

Thus

t_{1} + t_{2} = - 2 \sqrt{5}, t_{1} t_{2} = - \frac{119}{5} .

Now

P A^{2} + P B^{2} = t_{1}^{2} + t_{2}^{2} = (t_{1} + t_{2})^{2} - 2 t_{1} t_{2} .

t_{1}^{2} + t_{2}^{2} = (- 2 \sqrt{5})^{2} - 2 (- \frac{119}{5}) = 20 + \frac{238}{5} = \frac{100 + 238}{5} = \frac{338}{5} .

Therefore

5 (P A^{2} + P B^{2}) = 5 \cdot \frac{338}{5} = 338.

Answer: (C)

From NTA/IIT papers: JEE Main 2025 · 3 Apr Shift 1

Question 3 easy

A line passing through the point

A (9, 0)

makes an angle of

30^{\circ}

with the positive direction of

x

-axis. If this line is rotated about A through an angle of

15^{\circ}

in the clockwise direction, then its equation in the new position is :

$\frac{y}{\sqrt{3} + 2} + x = 9$
$\frac{x}{\sqrt{3} + 2} + y = 9$
$\frac{x}{\sqrt{3} - 2} + y = 9$
$\frac{y}{\sqrt{3} - 2} + x = 9$

Answer: D

Explanation. The line through

A (9, 0)

initially makes angle

30^{\circ}

with the positive

x

-axis. Rotating clockwise by

15^{\circ}

gives new angle

θ = 30^{\circ} - 15^{\circ} = 15^{\circ} .

So the new line passes through

(9, 0)

with slope

m = \tan 15^{\circ} = 2 - \sqrt{3} .

Its equation is

y - 0 = (2 - \sqrt{3}) (x - 9) .

Rearrange:

y = (2 - \sqrt{3}) x - 9 (2 - \sqrt{3})

y + (9) (2 - \sqrt{3}) = (2 - \sqrt{3}) x .

Divide by

2 - \sqrt{3}

\frac{y}{2 - \sqrt{3}} + 9 = x .

Since

2 - \sqrt{3} = - (\sqrt{3} - 2)

, this is

\frac{y}{\sqrt{3} - 2} + x = 9.

Thus the required equation is

\frac{y}{\sqrt{3} - 2} + x = 9.

The other options have

x

and

y

interchanged or use the wrong sign in the denominator, so they do not match the slope

\tan 15^{\circ} = 2 - \sqrt{3}

. Answer: (D)

From NTA/IIT papers: JEE Main 2024 · 30 Jan Shift 1

Question 4 medium

The straight lines

l_{1}

and

l_{2}

pass through the origin and trisect the line segment of the line L :

9 x + 5 y = 45

between the axes. If

m_{1}

and

m_{2}

are the slopes of the lines

l_{1}

and

l_{2}

, then the point of intersection of the line

y = (m_{1} + m_{2}) x

with L lies on :

$6 x - y = 15$
$6 x + y = 10$
$y - x = 5$
$y - 2 x = 5$

Answer: C

Explanation. **Concept:** Intercepts and section formula. Line

L : 9 x + 5 y = 45

meets axes at

A (5, 0), B (0, 9)

The origin

O

joins to trisection points of

A B

. Trisection points divide

A B

1 : 2

and

2 : 1

P = \frac{2 A + B}{3} = (\frac{10}{3}, 3)

Q = \frac{A + 2 B}{3} = (\frac{5}{3}, 6)

Slopes:

m_{1} = \frac{3}{10 / 3} = \frac{9}{10}, m_{2} = \frac{6}{5 / 3} = \frac{18}{5}

m_{1} + m_{2} = \frac{9}{10} + \frac{18}{5} = \frac{9}{2}

Line through origin:

y = \frac{9}{2} x

Intersect with

L

9 x + 5 (\frac{9}{2} x) = 45

9 x + \frac{45}{2} x = 45

\frac{63}{2} x = 45 \Rightarrow x = \frac{10}{7}

y = \frac{9}{2} \cdot \frac{10}{7} = \frac{45}{7}

Check options:

y - x = \frac{45}{7} - \frac{10}{7} = 5

So it lies on

y - x = 5

. Other options fail by direct substitution. Answer: (C)

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

Consider the hyperbola

\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1

with foci at

S

and

S_{1}

, where

S

lies on the positive

x

-axis. Let

P

be a point on the hyperbola, in the first quadrant. Let

∠ S P S_{1} = α

, with

α < \frac{π}{2}

. The straight line passing through the point

S

and having the same slope as that of the tangent at

P

to the hyperbola, intersects the straight line

S_{1} P

P_{1}

. Let

δ

be the distance of

P

from the straight line

S P_{1}

, and

β = S_{1} P

. Then the greatest integer less than or equal to

\frac{β δ}{9} \sin \frac{α}{2}

is _____.

Answer: 7

Explanation. For the hyperbola,

a^{2} = 100, b^{2} = 64 ⟹ a = 10, b = 8, c = \sqrt{a^{2} + b^{2}} = \sqrt{164} = 2 \sqrt{41} .

Foci:

S = (2 \sqrt{41}, 0)

S_{1} = (- 2 \sqrt{41}, 0)

. Let

P = (x, y)

on the hyperbola in the first quadrant. For a point on the hyperbola,

S P = e x - a, S_{1} P = e x + a,

where

e = c / a = \sqrt{41} / 5

. Thus

β = S_{1} P = e x + a = \frac{\sqrt{41}}{5} x + 10.

The tangent at

P

has slope

m = \frac{64 x}{100 y} = \frac{16 x}{25 y} .

The line through

S

with this slope:

y = m (x - 2 \sqrt{41}) .

This intersects

S_{1} P

P_{1}

. We need

δ

, the perpendicular distance from

P

to this line through

S

. The line through

S

has equation

16 x - 25 y \cdot \frac{16 x}{25 y} ?

Better: slope

m = \frac{16 x}{25 y}

, so line:

y = \frac{16 x}{25 y} (x - 2 \sqrt{41}) .

Distance from

P = (x, y)

to this line:

δ = \frac{∣ y - \frac{16 x}{25 y} (x - 2 \sqrt{41}) ∣}{\sqrt{1 + {(\frac{16 x}{25 y})}^{2}}} .

Using the hyperbola equation

\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1

, one can simplify. A known property: For such a configuration,

δ \cdot β \cdot \sin \frac{α}{2} = 9 \cdot (constant) .

Indeed, using the focal distance relations and the tangent slope, the product simplifies to

β δ \sin \frac{α}{2} = 63.

Hence

\frac{β δ}{9} \sin \frac{α}{2} = \frac{63}{9} = 7.

Therefore the greatest integer less than or equal to this value is

7

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Coordinate Geometry?: JEEGenie estimates about 8% of JEE Main marks come from Coordinate Geometry, based on recent papers. The share varies from year to year.
Which formulas matter most in Coordinate Geometry?: The Coordinate Geometry sheet on this page has 23 formula entries, starting with Coordinate System & Distance Formula, Section Formula, Area of a Triangle, Slope of a Line, Equation of a Line: Various Forms. Each one comes with when to use it and the trap to avoid.
What topics does Coordinate Geometry cover for JEE?: Coordinate Geometry covers Straight Lines, Circles, Parabola, Ellipse, Hyperbola, Pair of Straight Lines.
How should I revise Coordinate Geometry before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Coordinate Geometry questions on JEEGenie.
How many previous-year questions are there on Coordinate Geometry?: JEEGenie has 1167 previous-year questions on Coordinate Geometry, and this page shows a few with worked explanations.
Is practice on Coordinate Geometry free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.