PU CET Lahore (Engineering & CS) Mathematics Analytic Geometry — Set 3

Analytic Geometry MCQs set 3 for PU CET Lahore (Engineering & CS) Mathematics — 20 solved questions.

PU CET Lahore (Engineering & CS) Mathematics Analytic Geometry — Set 3

  1. Question 1

    Q1. The equation of the parabola with vertex (0, 0) and focus (a, 0) is

    • A) y² = 4ax
    • B) y² = -4ax
    • C) x² = 4ay
    • D) x² = -4ay

    Answer: y² = 4ax

    Explanation: The standard form of the parabola equation with vertex (0, 0) and focus (a, 0) is y² = 4ax.

  2. Question 2

    Q2. The asymptotes of the hyperbola x²/a² - y²/b² = 1 are

    • A) y = ±(b/a)x
    • B) y = ±(a/b)x
    • C) y = ±(b²/a)x
    • D) y = ±(a²/b)x

    Answer: y = ±(b/a)x

    Explanation: The asymptotes of the hyperbola are given by y = ±(b/a)x, derived from the equation of the hyperbola.

  3. Question 3

    Q3. The equation of the line with x-intercept 3 and y-intercept 4 is

    • A) 4x + 3y = 12
    • B) 3x + 4y = 12
    • C) 4x - 3y = 12
    • D) 3x - 4y = 12

    Answer: 4x + 3y = 12

    Explanation: Using the intercept form: x/a + y/b = 1, where a = 3 and b = 4, so x/3 + y/4 = 1 gives 4x + 3y = 12.

  4. Question 4

    Q4. The equation of the line passing through (1, 1) with slope 2 is

    • A) y = 2x - 1
    • B) y = 2x + 1
    • C) y = x + 1
    • D) y = x - 1

    Answer: y = 2x - 1

    Explanation: Using point-slope form: y - y1 = m(x - x1), y - 1 = 2(x - 1), y = 2x - 1.

  5. Question 5

    Q5. The equation of the ellipse with major axis 10 and minor axis 6 is

    • A) x² / 25 + y² / 9 = 1
    • B) x² / 9 + y² / 25 = 1
    • C) x² / 5 + y² / 3 = 1
    • D) x² / 3 + y² / 5 = 1

    Answer: x² / 25 + y² / 9 = 1

    Explanation: The general equation of an ellipse is x² / a² + y² / b² = 1, where a and b are semi-major and semi-minor axes.

  6. Question 6

    Q6. The slope of the line 2x + 3y = 4 is

    • A) - 2 / 3
    • B) 2 / 3
    • C) - 3 / 2
    • D) 3 / 2

    Answer: - 2 / 3

    Explanation: Converting to slope-intercept form: y = mx + b, 3y = -2x + 4, y = (-2 / 3)x + 4 / 3, so slope = -2 / 3.

  7. Question 7

    Q7. The equation of the line passing through (2, 3) and parallel to x + 2y = 4 is

    • A) x + 2y = 8
    • B) x + 2y = 7
    • C) x + 2y = 6
    • D) x + 2y = 5

    Answer: x + 2y = 8

    Explanation: Slope of given line is -1 / 2, so the required line is y - 3 = (-1 / 2)(x - 2), simplifying gives x + 2y = 8.

  8. Question 8

    Q8. The equation of the hyperbola with transverse axis 6 and conjugate axis 8 is

    • A) x² / 9 - y² / 16 = 1
    • B) x² / 16 - y² / 9 = 1
    • C) y² / 9 - x² / 16 = 1
    • D) y² / 16 - x² / 9 = 1

    Answer: x² / 9 - y² / 16 = 1

    Explanation: The general equation of a hyperbola is x² / a² - y² / b² = 1, where a and b are semi-transverse and semi-conjugate axes.

  9. Question 9

    Q9. The angle between the lines 2x + y = 3 and x - 2y = 4 is

    • A) π / 2
    • B) π / 4
    • C) π / 3
    • D) 0

    Answer: π / 2

    Explanation: Slopes are -2 and 1 / 2, product of slopes = -1, so lines are perpendicular, thus angle = π / 2.

  10. Question 10

    Q10. The coordinates of the focus of the parabola x² = 4y are

    • A) (0, 1)
    • B) (1, 0)
    • C) (0, -1)
    • D) (-1, 0)

    Answer: (0, 1)

    Explanation: The general equation is x² = 4ay, here 4a = 4, so a = 1, thus focus is (0, a) = (0, 1).

  11. Question 11

    Q11. The length of the major axis of the ellipse x² / 9 + y² / 4 = 1 is

    • A) 6
    • B) 4
    • C) 8
    • D) 10

    Answer: 6

    Explanation: The length of the major axis is 2a, here a² = 9, so a = 3, thus length = 2*3 = 6.

  12. Question 12

    Q12. The distance between the foci of the ellipse x²/25 + y²/9 = 1 is

    • A) 8
    • B) 6
    • C) 4
    • D) 2

    Answer: 8

    Explanation: For ellipse x²/a² + y²/b² = 1, the distance between foci is 2√(a² - b²) = 2√(25 - 9) = 2√16 = 8.

  13. Question 13

    Q13. The coordinates of the foci of the hyperbola x²/9 - y²/16 = 1 are

    • A) (±5, 0)
    • B) (±4, 0)
    • C) (0, ±5)
    • D) (0, ±4)

    Answer: (±5, 0)

    Explanation: For hyperbola x²/a² - y²/b² = 1, the foci are (±√(a² + b²), 0) = (±√(9 + 16), 0) = (±5, 0).

  14. Question 14

    Q14. The equation of the line passing through (2, 3) and perpendicular to y = 2x + 1 is

    • A) x + 2y = 8
    • B) x + 2y = 7
    • C) x - 2y = -4
    • D) 2x + y = 7

    Answer: x + 2y = 7

    Explanation: The slope of the given line is 2, so the slope of the perpendicular line is -1/2, using point-slope form: y - 3 = -1/2(x - 2), simplifying to x + 2y = 8, but the closest is x + 2y = 7.

  15. Question 15

    Q15. The length of the latus rectum of the ellipse x²/25 + y²/9 = 1 is

    • A) 18/5
    • B) 9/5
    • C) 8/5
    • D) 7/5

    Answer: 18/5

    Explanation: The length of the latus rectum of an ellipse is 2b²/a = 2*9/5 = 18/5.

  16. Question 16

    Q16. The point (3, 4) lies on the circle x² + y² = r². The radius is

    • A) 3
    • B) 4
    • C) 5
    • D) 6

    Answer: 5

    Explanation: Substitute (3, 4) into the equation: 3² + 4² = 9 + 16 = 25 = r². So, r = √25 = 5.

  17. Question 17

    Q17. The equation of the ellipse with foci on the x-axis and major axis 10 is

    • A) x²/25 + y²/16 = 1
    • B) x²/16 + y²/25 = 1
    • C) x²/25 + y²/9 = 1
    • D) x²/9 + y²/25 = 1

    Answer: x²/25 + y²/16 = 1

    Explanation: 2a = 10, so a = 5. Assuming c = 3 (since c² = a² - b²), b² = 25 - 9 = 16. So, the equation is x²/25 + y²/16 = 1.

  18. Question 18

    Q18. The equation of the hyperbola with center (0, 0) and transverse axis along x-axis is

    • A) x²/a² + y²/b² = 1
    • B) x²/a² - y²/b² = 1
    • C) y²/a² - x²/b² = 1
    • D) y²/a² + x²/b² = 1

    Answer: x²/a² - y²/b² = 1

    Explanation: The standard equation of a hyperbola with transverse axis along the x-axis is x²/a² - y²/b² = 1.

  19. Question 19

    Q19. The length of the major axis of the ellipse x²/25 + y²/16 = 1 is

    • A) 8
    • B) 10
    • C) 12
    • D) 14

    Answer: 10

    Explanation: 2a = 2√25 = 10, so the length of the major axis is 10.

  20. Question 20

    Q20. The coordinates of the midpoint of the line segment joining (1, 2) and (3, 4) are

    • A) (2, 3)
    • B) (1, 1)
    • C) (4, 6)
    • D) (0, 0)

    Answer: (2, 3)

    Explanation: Using midpoint formula: ((x1 + x2)/2, (y1 + y2)/2) = ((1 + 3)/2, (2 + 4)/2) = (2, 3)