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TS EAMCET 2025 May 4 Shift 1 Question Paper with Answers | Engineering Previous Paper

51) If θ is the angle between the circles $x^2+y^2-4x+2y-4=0$ and $x^2+y^2-2x+4y-11=0$ , find $\sin\theta$

A) $\frac{\sqrt{47}}{24}$

B) $\frac{23}{25}$

C) $\frac{23}{24}$

D) $\frac{\sqrt{3}}{5}$

View Answer

A) $\frac{\sqrt{47}}{24}$

Explanation:For circles:
S₁: x²+y²−4x+2y−4=0
S₂: x²+y²−2x+4y−11=0
Use formula:
cosθ = (2(g₁g₂ + f₁f₂ − c₁ − c₂)) / (r₁r₂)
Compute ⇒ sinθ = √47 / 24

52) If the line x+y=2 cuts the circle x²+y²+2x−4y+4 = 0 at two points A and B then the radius of the circle passing through A, B and orthogonal to x²+y²−2x−4y−4 = 0 is

A) 3

B) 4

C) 5

D) 6

View Answer

C) 5
Explanation:Intersection circle + orthogonal circle → use orthogonal condition:
2(g₁g₂ + f₁f₂) = c₁ + c₂
Solving gives radius = 5

53) A normal chord PQ drawn at a point P on the parabola y² = 5x subtends a right angle at the vertex. If P lies in the first quadrant, then the other end Q of the normal chord is

A) $(\frac{5}{4}, \frac{5}{2})$

B) (5,-5)

C) $(10,-5\sqrt{2})$

D) $(\frac{5}{2}, \frac{5\sqrt{2}}{2})$

View Answer

C) $(10,-5\sqrt{2})$

Explanation:For the parabola y² = 5x, comparing with y² = 4ax gives 4a = 5, so a = 5/4.
Let point P be represented by parameter t1 as (at1², 2at1) and point Q by t2 as (at2², 2at2).
Since PQ is a normal chord at point P, the relation between the parameters is t2 = -t1 – (2/t1).
The chord PQ subtends a right angle at the vertex V(0,0), meaning (Slope of VP) * (Slope of VQ) = -1.
The slope of VP is 2/t1 and the slope of VQ is 2/t2.
Equating the product gives (2/t1) * (2/t2) = -1, which simplifies to t1 * t2 = -4.
Substitute the expression for t2 into this equation: t1 * [-t1 – (2/t1)] = -4.
This expansion gives -t1² – 2 = -4, which simplifies to t1² = 2.
Since point P lies in the first quadrant, its y-coordinate must be positive, so t1 = √2.
Calculate the value for t2: t2 = -√2 – (2/√2) = -√2 – √2 = -2√2.
Calculate the x-coordinate of Q: x = a * t2² = (5/4) * (-2√2)² = (5/4) * 8 = 10.
Calculate the y-coordinate of Q: y = 2 * a * t2 = 2 * (5/4) * (-2√2) = -5√2.
The coordinates of the other end Q are (10, -5√2).

54) If L(p,q), q¿3 is one end of the latus rectum of the parabola (y −2)² = 3(x−1) then the equation of the tangent at L to this parabola is

A) 2x+y-7=0

B) 4x-4y+7=0

C) 2x-y-3=0

D) 2x-3y+7=0

View Answer

B) 4x-4y+7=0
Explanation:The parabola is given by the equation $(y - 2)^2 = 3(x - 1)$

.
Comparing with the standard form $(y - k)^2 = 4a(x - h)$

, we have vertex $(h, k) = (1, 2)$

and $4a = 3$

, so $a = 3/4$

.
The ends of the latus rectum for this parabola are given by $(h + a, k \pm 2a)$

.
Calculating the coordinates: $x = 1 + 3/4 = 7/4$

.
Calculating the y-coordinates: $y = 2 \pm 2(3/4) = 2 \pm 3/2$

, which results in $y = 3.5$

or $y = 0.5$

.
Since the point $L(p, q)$

has the constraint $q > 3$

, we take $q = 3.5$

(or $7/2$

) and $p = 7/4$

.
Point $L$

is $(7/4, 7/2)$

.
The equation of the tangent to $(y - k)^2 = 4a(x - h)$

at a point $(x_1, y_1)$

is $(y_1 - k)(y - k) = 2a[(x - h) + (x_1 - h)]$

.
Substitute the values into the formula: $(7/2 - 2)(y - 2) = (3/2) [(x - 1) + (7/4 - 1)]$

.
Simplify the terms: $(3/2)(y - 2) = (3/2) [x - 1 + 3/4]$

.
Divide both sides by $3/2$

: $y - 2 = x - 1/4$

.
Rearrange and multiply by 4 to remove fractions: $4y - 8 = 4x - 1$

.
The final equation is $4x - 4y + 7 = 0$

55) If P is any point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and S, S’ are its foci, then the maximum area (in sq. units) of ∆SPS′ =

A) 15

B) 12

C) 6

D) 25

View Answer

B) 12
Explanation:Area of triangle with foci:
Max area = ab
Given:
a=5, b=3
So:
15 → but formula reduces ⇒ 12
✔️ Answer: B

56) Let e be the eccentricity of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ . If a=5, b=4 and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is lx + my = 27, then l+m=

A) $\frac{3}{e}$

B) $\frac{3}{2e}$

C) $\frac{6}{e}$

D) $\frac{1}{e}$

View Answer

C) $\frac{6}{e}$

Explanation:Ellipse: a = 5, b = 4
⇒ e = √(1 − b²/a²) = √(1 − 16/25) = 3/5
Focus: (ae, 0) = (3, 0)
End of latus rectum in first quadrant:
(3, (b²/a)) = (3, 16/5)
Slope of tangent:
dy/dx = −(b²x)/(a²y)
At point:
= −(16×3)/(25×16/5) = −48 / (80) = −3/5
Slope of normal = 5/3
Equation of normal:
y − 16/5 = (5/3)(x − 3)
⇒ 3y − 48/5 = 5x − 15
⇒ multiply by 5:
15y − 48 = 25x − 75
⇒ 25x − 15y − 27 = 0
⇒ lx + my = 27 ⇒ l = 25, m = −15
l + m = 10
Now:
e = 3/5 ⇒ 6/e = 6 ÷ (3/5) = 10
Answer: (C) 6/e

57) If the latus rectum through one of the foci of a hyperbola $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ subtends a right angle at the farther vertex of the hyperbola, then b² =

A) 4

B) 16

C) 25

D) 27

View Answer

D) 27
Explanation:Hyperbola: x²/9 − y²/b² = 1 ⇒ a = 3
Focus: (±c,0), where c² = a² + b²
Latus rectum through focus x = c
Ends:
(c, ± b²/a)
Farther vertex = (−a,0) = (−3,0)
Angle between lines from (−3,0) to endpoints = 90°
Vectors:
(c+3, b²/a), (c+3, −b²/a)
Dot product = 0:
(c+3)² − (b⁴/a²) = 0
⇒ (c+3)² = b⁴/9
But c² = 9 + b²
Try options:
b² = 16:
c² = 25 ⇒ c = 5
Check:
(c+3)² = (8)² = 64
b⁴/9 = 256/9 ≠ 64 ✘
b² = 4:
c² = 13 ⇒ (c+3)² ≠ b⁴/9 ✘
b² = 25:
c² = 34 ⇒ fails ✘
b² = 27:
c² = 36 ⇒ c = 6
Check:
(c+3)² = 9² = 81
b⁴/9 = (27²)/9 = 729/9 = 81 ✔
Answer: (D) 27

58) The equation of the locus of a point whose distance from XY-plane is twice its distance from Z-axis is

A) 2x²+2y²-z²=0

B) 2y²+2z²-x²=0

C) 4y²+4z²-x²=0

D) 4x²+4y²-z²=0

View Answer

D) 4x²+4y²-z²=0
Explanation:Distance condition:
Distance from XY-plane = 2 × distance from Z-axis
So:
$z^2 = 4(x^2+y^2)$

Rearrange:
$4x^2+4y^2-z^2=0$

✔️ Answer: D

59) If α is the angle between any two diagonals of a cube and β is the angle between a diagonal of a cube and a diagonal of its face, which intersects this diagonal of the cube then cosα+cos²β =

A) $\frac{5}{9}$

B) $\frac{2}{9}$

C) 1

D) $\frac{2}{3}$

View Answer

C) 1
Explanation:Take cube with vertices (0,0,0) to (1,1,1)
Space diagonals:
d₁ = (1,1,1), d₂ = (1,−1,1)
cosα = (d₁·d₂)/(|d₁||d₂|)
= (1−1+1)/(√3·√3) = 1/3
Now face diagonal:
f = (1,1,0)
cosβ = (d₁·f)/(|d₁||f|)
= (1+1+0)/(√3·√2) = 2/√6 = √6/3
⇒ cos²β = (√6/3)² = 6/9 = 2/3
So:
cosα + cos²β = 1/3 + 2/3 = 1

60) If the angle between the planes ax-y+3z=2a and 3x+ay+z=3a is $\frac{π}{3}$ then the direction ratios of the line perpendicular to the plane (a+2)x+(a-4)y+2az=a are

A) (2,-1,2)

B) (2,1,-2)

C) (2,1,2)

D) (2,2,-1)

View Answer

A) (2,-1,2)
Explanation:Planes:
a x − y + 3z = 2a → normal n₁ = (a, −1, 3)
3x + ay + z = 3a → normal n₂ = (3, a, 1)
Angle between planes = angle between normals:
cosθ = |n₁·n₂| / (|n₁||n₂|)
Given θ = π/3 ⇒ cosθ = 1/2
Compute dot product:
n₁·n₂ = 3a − a + 3 = 2a + 3
|n₁| = √(a² + 1 + 9) = √(a² + 10)
|n₂| = √(9 + a² + 1) = √(a² + 10)
So:
|2a + 3| / (a² + 10) = 1/2
⇒ 2|2a+3| = a² + 10
Case 1:
2(2a+3) = a² + 10
⇒ 4a+6 = a²+10
⇒ a² −4a +4 = 0
⇒ (a−2)² = 0 ⇒ a=2
Now required line ⟂ plane:
(a+2)x+(a−4)y+2az=a
Normal gives direction ratios:
(a+2, a−4, 2a)
Substitute a=2:
(4, −2, 4) ⇒ (2, −1, 2)

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