COMEDK UGET 2025 Coordinate Geometry Practice Questions with Solutions

COMEDK UGET 2025 Coordinate Geometry Practice Questions with Solutions: The chapter on Coordinate Geometry covers topics such as Straight Lines, Conic Sections, and Three-dimensional Geometry. If you are preparing for the COMEDK UGET 2025 exam , you must prepare Coordinate Geometry diligently as it carries around 12-13 questions. You should solve COMEDK UGET 2025 Coordinate Geometry practice questions with solutions from previous year papers to ensure a comprehensive preparation for this section. The majority of the topics will be asked from COMEDK Coordinate Geometry sample questions. As per the latest exam analysis, Coordinate Geometry is expected to have approximate 20% weighatge, which can significantly affect your overall score. In this article, we have provided some of the most important COMEDK Coordinate practice questions and expected weightage for the upcoming entrance exam.

Also Check - Do or Die Chapters for COMEDK UGET 2025 Mathematics

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COMEDK UGET 2025 Coordinate Geometry Important Topics

The authorities have provided the detailed COMEDK UGET 2025 Mathematics syllabus to help students prepare for the upcoming exam. COMEDK Coordinate Geometry question paper covers mainly three topics: Straight Lines, Conic Sections, and Three-Dimensional Geometry. In the following table, we have listed the section-wise syllabus here:

COMEDK UGET 2025 Coordinate Geometry Expected Weightage

Coordinate Geometry carries some of the highest weightage, around 20%. Around 12-13questios are asked from COMEDK Coordinate Geometry sample questions. In the following table, we have detailed Coordinate Geometry expected weighatge in COMEDK UGET 2025:

COMEDK UGET 2025 Coordinate Geometry Practice Questions with Solutions

COMEDK UGET 2025 Coordinate Geometry practice questions with solutions are provided below. You should solve these questions to practice for the exam in a diligent manner.

Q1. Let ABC be a triangle with equations of its sides AB, BC, CA respectively are x - 2 = 0, y - 5 = 0, and 5x + 2y - 10 = 0. Then the orthocentre of the triangle lies on the line is?

a. 4x + y = 13

b. 3x + y = 1

c. x - y = 0

d. X - 2y = 0

Ans. a. 4x + y = 13

Solution: We are given that the sides of triangle ABC are:

AB: x - 2 = 0

BC: y - 5 = 0

CA: 5x + 2y - 10 = 0

Let's break it down step by step.

Determine the vertices by finding the intersections of the lines:

Intersection of x - 2 = 0 and y - 5 = 0

x = 2, y = 5

So one vertex is B(2, 5)

Intersection of x - 2 = 0 and 5x + 2y - 10 = 0

Substitute x = 2 into 5x + 2y - 10 = 0

5(2) + 2y - 10 = 0; 2y = 0; y = 0

Thus, the corresponding vertex is A(2,0)

Intersection of y - 5 = 0 and 5x + 2y - 10 = 0

Substitute y = 5 into 5x + 2y - 10 = 0

5x + 2(5) - 10 = 0; 5x = 0; x = 0

Therefore, the third vertex is C(0,5)

Identify the right angle:

Notice that line x - 2 = 0, is vertical and line y - 5 = 0, is horizontal. These lines are perpendicular, so the angle at their intersection, point B(2,5), is 90°.

Find the orthocentre:

In a right-angled triangle, the orthocentre is located at the vertex where the right angle occurs. Since the right angle is at B(2,5), the orthocentre is also B(2,5)

Check which of the given lines passes through the orthocentre B(2,5)

Option A: 3x + y = 1

Substitute x = 2 and y = 5:

3(2) + 5 = 6 + 5 = 11 (≠ 1)

Thus, the orthocentre (2,5) lies on the line 4x + y = 13, which is option a.

Q2. The equation of a circle passing through the origin is x ² + y ² - 6x + 2y = 0. The equation of one of its diameters is:

a. 3x - y = 0

b. x + 3y = 0

c. x - 3y = 0

d. x + y = 0

Ans. b. x + 3y = 0

Solution: To determine the equation of one of the diameters of the given circle, we need to first rewrite the equation of the circle in its standard form. The given equation is:

x ² + y ² - 6x + 2y = 0

We will complete the square for both the x and y terms:

Starting with the x terms:

x ² - 6x

Add and subtract (6/2) ² = 9

x ² - 6x + 9 - 9 = (x - 3) ² - 9

Next, for the y terms:

y ² + 2y

Add and subtract (2/2) ² = 1

y ² + 2y + 1 - 1 = (y+1) ² - 1

Now substitute these results back into the original equation:

x ² - 6x + y ² +2y = 0

Becomes:

[(x-3) ² - 9] + [(y+1) ² - 1] = 0

Simplify it,

(x-3) ² + (y+1) ² - 10 = 0

This is equivalent to:

(x-3) ² + (y+1) ² = 10

This represents a circle with center at (3, -1) and radius √10

The equation of a diameter of the circle must pass through the center, (3, -1), and the origin, (0,0). We can find the equation of the line passing through these two points using the point-slope form.

The slope of the line passing through (3, -1) and (0,0) is:

Slope [0-(-1)]/0-3 = 1/-3 = -1/3

Using the point-slope form of the line equation:

y - y ₁ = m(x - x ₁ )

Where, (x ₁ ,y ₁ ) = (3,-1) and m = -1/3, we get,

y + 1 = -1/3 (x - 3)

Multiply through by 3 to clear the fraction:

3(y+1) = -(x-3)

Simplify the equation:

3y + 3 = -x + 3

Rearrange to get the standard form of the line equation:

x + 3y = 0

Therefore, the equation of one of its diameters is:

Option D.

Q3. In the parabola y ² = 4ax, the length of the latus rectum is 6 units and there is a chord passing through its vertex and the negative end of the latus rectum. Then the equation of the chord is?

a. x + 2y = 0

b. x -2y = 0

c. 2x + y = 0

d. 2x - y = 0

Ans. c. 2x + y = 0

Solution: The length of the latus rectum of the parabola y ² = 4ax is 4a

Given that the length of the latus rectum is 6 units, we have 4a = 6. Therefore, a = 3/2

The equation of the parabola becomes y ² = 6x

The vertex of the parabola is at the origin (0,0), and the negative end of the latus rectum is at the point (-a, 2a), which is (-3/2, 3) in this case. The slope of the chord passing through the vertex and the negative end of the latus rectum is (3-0)/-3/2-0 = -2.

The equation of the chord in point-slope form is y - 0 = -2(x-0), which simplifies to y = -2x

Therefore, the equation of the chord is 2x + y = 0

Q4. If the length of the major axis of an ellipse is 3 times the length of the minor axis, then its eccentricity is

a. 1/√2

b. 2/√3

c. 1/√3

d. 2√2/3

Ans. 2√2/3

Solution: To find the eccentricity of the ellipse, we start with the given information that the length of the major axis of an ellipse is 3 times the length of the minor axis. First, let's define the standard notation for an ellipse and apply the given information.

The standard form of the ellipse with the major and minor axes along the x-axis and y-axis respectively is

x ² /a ² + y ² /b ² = 1.

where a is the semi-major axis and b is the semi-minor axis. The length of the major axis is 2a and the length of the minor axis is 2b. Given that the length of the major axis is 3 times the length of the minor axis, we have: 2a = 3(2b)

Or,

a = 3b

The eccentricity e of an ellipse is given by the formula:

e = √(1-b ² /a ² )

Substitute a = 3b into the eccentricity formula:

e = √[1-b ² /(3b) ² ]

Simplifying the equation,

e = √8/2

Or, e = 2√2/3

Q5. If the distance between the foci and the distance between the two directrices are in the ratio 3:2 for a hyperbola x ² /a ² - y ² /b ² = 1; then a:b is:

a. √2:1

b. 1:2

c. √3:√2

d. 2:1

Ans. a. √2:1

Solution: Let's work through the problem step-by-step.

For the hyperbola

x ² /a ² - y ² /b ² = 1

the foci are located at

± c with c ² = a ² + b ²

Thus, the distance between the foci is 2c.

The directrices of this hyperbola are given by,

x = ± a/e,

Where the eccentricity is

e = c/a.

Therefore, the distance between the two directrices is

2(a/e)

According to the problem, the ratio of the distance between the foci to the distance between the directrices is

2c/2(a/e) = ce/a = 3/2

Since the eccentricity is

e = c/a.

Multiplying by e, e ² = ce/a

e ² = 3/2

We also know that for this hyperbola,

e ² = 1+b ² /a ²

Setting these equal gives:

1 + b ² /a ² = 3/2

Subtract 1 from both sides

b ² /a ² = 3/2 - 1 = 1/2

Taking square roots on both sides, we find:

b/a = 1/√2,

which can be rewritten as

b:a = √2:1

Coordinate Geometry covers a significant part of the Mathematics syllabus in COMEDK exam, thus making it one of the most important chapters. You should focus on important chapters from the syllabus, and solve as many COMEDK UGET 2025 Coordinate Geometry practice questions as possible.

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