MATHEMATICS (Analytical Geometry) | FYUG Even Semester Exam, 2025

Question: Find the transformed equation of the line x cos α + y sin α = p, when the axes are rotated through an angle α. [2 Marks]

Solution:

Let the axes be rotated through an angle α. The transformation equations are:

• x = x' cos α - y' sin α
• y = x' sin α + y' cos α

Substituting these into the given equation x cos α + y sin α = p:

(x' cos α - y' sin α) cos α + (x' sin α + y' cos α) sin α = p

x' cos² α - y' sin α cos α + x' sin² α + y' sin α cos α = p

x' (cos² α + sin² α) = p

x' = p

Final Answer: The transformed equation is x' = p.

Question: Prove that 6x² - 5xy - 6y² + 14x + 5y + 4 = 0 represents a pair of perpendicular lines. [2 Marks]

Solution:

For a second-degree equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 to represent perpendicular lines, the sum of the coefficients of x² and y² must be zero.

Condition: a + b = 0

From the given equation:

• a = 6
• b = -6

Calculation: a + b = 6 + (-6) = 0.

Since a + b = 0, the lines represented by the equation are perpendicular to each other.

Question: The origin is shifted to (3,1) and the axes rotated through θ = tan⁻¹(3/4). Find coordinates of (5,-2) in the new system. [2 Marks]

Solution:

1. Translation: Shift origin to (h, k) = (3, 1). New coordinates (x₁, y₁):

• x₁ = x - h = 5 - 3 = 2
• y₁ = y - k = -2 - 1 = -3

2. Rotation: Rotate through θ where tan θ = 3/4. This implies sin θ = 3/5 and cos θ = 4/5.

New coordinates (x', y'):

• x' = x₁ cos θ + y₁ sin θ = 2(4/5) + (-3)(3/5) = 8/5 - 9/5 = -1/5
• y' = -x₁ sin θ + y₁ cos θ = -2(3/5) + (-3)(4/5) = -6/5 - 12/5 = -18/5

Final Answer: (-0.2, -3.6)

Question: Prove that ax² + 2hxy + by² = 0 always represents a pair of straight lines passing through the origin. [5 Marks]

Solution:

The given equation is a homogeneous equation of second degree in x and y. Divide by x² (assuming x ≠ 0):

b(y/x)² + 2h(y/x) + a = 0

Let m = y/x be the slope of the lines. Then:

bm² + 2hm + a = 0

This is a quadratic equation in m, giving two values m₁ and m₂. If these roots are real (i.e., h² ≥ ab), the equation represents two lines:

• y = m₁x
• y = m₂x

Both equations satisfy (0,0), proving they pass through the origin.

Question: Prove that the lines ax² + 2hxy + by² = 0 and a²x² + 2h(a+b)xy + b²y² = 0 have the same bisectors. [5 Marks]

Solution:

The equation of the bisectors of the angles between the lines ax² + 2hxy + by² = 0 is:

(x² - y²) / (a - b) = xy / h

Now consider the second equation: A = a², H = h(a+b), B = b².

The bisector equation for this is: (x² - y²) / (A - B) = xy / H

Substituting values: (x² - y²) / (a² - b²) = xy / [h(a+b)]

(x² - y²) / [(a - b)(a + b)] = xy / [h(a+b)]

Canceling (a+b) from both denominators: (x² - y²) / (a - b) = xy / h.

Both result in the same equation, hence they share the same bisectors.

Question: Find the value of k for which x² + y² + 5x + 3y + 7 = 0 and x² + y² - 8x + 6y + k = 0 are orthogonal. [2 Marks]

Solution:

Condition for orthogonality of two circles: 2g₁g₂ + 2f₁f₂ = c₁ + c₂

• Circle 1: g₁ = 5/2, f₁ = 3/2, c₁ = 7
• Circle 2: g₂ = -4, f₂ = 3, c₂ = k

2(5/2)(-4) + 2(3/2)(3) = 7 + k

-20 + 9 = 7 + k

-11 = 7 + k => k = -18

Final Answer: k = -18

Question: Prove that the locus of the point of intersection of perpendicular tangents to a parabola is the directrix. [5 Marks]

Solution:

Let the parabola be y² = 4ax. The equation of a tangent with slope 'm' is:

y = mx + a/m

If the tangent passes through (h, k), then: k = mh + a/m => m²h - mk + a = 0.

This quadratic in 'm' gives slopes m₁ and m₂ of the two tangents. If tangents are perpendicular, m₁m₂ = -1.

From the quadratic equation, product of roots m₁m₂ = a/h.

So, a/h = -1 => h = -a.

Replacing h with x, the locus is x = -a, which is the equation of the directrix.

Question: Prove that the sum of the reciprocals of two perpendicular focal chords of a conic is constant. [5 Marks]

Solution:

Question: Define shortest distance between two skew lines. What is the shortest distance between two intersecting lines? [2 Marks]

Solution:

The Shortest Distance (SD) between two skew lines is the length of the common perpendicular segment intercepted between the two lines.

For intersecting lines, the shortest distance is zero because they meet at a point.