Q5. Find the focus, axis, directrix and latus rectum of x2 = −16y (NCERT Exercise 10.2, Q4).
4a = 16, a = 4, opening downwards: focus (0, −4), axis the y-axis, directrix y = 4, latus rectum 16.
Q6. Find the focus, axis, directrix and latus rectum of y2 = 10x (NCERT Exercise 10.2, Q5).
4a = 10, a = 5/2: focus (5/2, 0), axis the x-axis, directrix x = −5/2, latus rectum 10.
Q7. Find the focus, axis, directrix and latus rectum of x2 = −9y (NCERT Exercise 10.2, Q6).
4a = 9, a = 9/4, opening downwards: focus (0, −9/4), axis the y-axis, directrix y = 9/4, latus rectum 9.
Q8. Find the parabola with focus (0, −3) and directrix y = 3 (NCERT Exercise 10.2, Q8).
The vertex is the origin, the focus is below it and a = 3: x2 = −12y.
Q9. Find the parabola with vertex (0, 0) and focus (3, 0) (NCERT Exercise 10.2, Q9).
a = 3, opening right: y2 = 12x.
Q10. Find the parabola with vertex (0, 0), symmetric about the y-axis, passing through (5, 2) (NCERT Exercise 10.2, Q12).
The point is above the x-axis, so x2 = 4ay: 25 = 8a, a = 25/8. So x2 = (25/2)y, that is 2x2 = 25y.
Q11. On the parabola y2 = 12x, find the point whose distance from the focus is 7.
PF = x + a = x + 3 = 7, so x = 4 and y2 = 48, y = ±4√3. The points are (4, 4√3) and (4, −4√3).
Q12. Show that the point (at2, 2at) lies on y2 = 4ax for every t, and find t for the end (a, 2a) of the latus rectum.
(2at)2 = 4a2t2 = 4a(at2). For (a, 2a), 2at = 2a gives t = 1 (and then at2 = a as well).
Q13. The focus of a parabolic mirror is 5 cm from its vertex. If the mirror is 45 cm deep, find the width AB across its rim (NCERT Miscellaneous Example 17, page 202).
With the vertex at the origin and the axis along the x-axis, a = 5 and y2 = 20x. At the rim x = 45, so y2 = 900 and y = ±30: AB = 60 cm.
Q14. A beam is supported at its ends by supports 12 m apart. The load at its centre causes a deflection of 3 cm there, and the deflected beam is a parabola. How far from the centre is the deflection 1 cm (NCERT Miscellaneous Example 18, pages 202–203)?
Put the vertex at the lowest point with the axis vertical: x2 = 4ay, through (6, 3/100) in metres, so 36 = 4a × 3/100 and a = 300. A deflection of 1 cm is 2 cm above the lowest point, so x2 = 4 × 300 × 2/100 = 24 and x = 2√6 ≈ 4.90 m from the centre.
Q15. A parabolic reflector is 20 cm across and 5 cm deep. Find its focus (NCERT Miscellaneous Exercise, Q1).
y2 = 4ax passes through (5, 10): 100 = 20a, a = 5. The focus is (5, 0), 5 cm from the vertex, at the centre of the rim.
Q16. An arch is a parabola with a vertical axis, 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex (NCERT Miscellaneous Exercise, Q2)?
With the vertex at the top, x2 = −4ay passes through (2.5, −10): 6.25 = 40a, a = 5/32. At y = −2, x2 = 4 × (5/32) × 2 = 5/4, so x = √5/2 and the width is √5 ≈ 2.24 m.
Q17. A suspension-bridge cable hangs as a parabola over a horizontal roadway 100 m long. The vertical wires are 30 m long at the ends and 6 m at the middle. Find the length of the wire 18 m from the middle (NCERT Miscellaneous Exercise, Q3).
Put the lowest point of the cable at (0, 6) above the road: x2 = 4a(y − 6) through (50, 30) gives 4a = 2500/24. At x = 18, y − 6 = 324 × 24/2500 ≈ 3.11, so the wire is about 9.11 m long.
Q18. Find the area of the triangle formed by joining the vertex of x2 = 12y to the ends of its latus rectum (NCERT Miscellaneous Exercise, Q6).
a = 3, so the latus rectum joins (−6, 3) and (6, 3). The triangle has base 12 and height 3: area 18 square units.
Q19. An equilateral triangle is inscribed in the parabola y2 = 4ax with one vertex at the vertex of the parabola. Find the length of its side (NCERT Miscellaneous Exercise, Q8).
Let O be the vertex. For P(x, y) on the parabola, OP = x + 4ax, which strictly increases with x ≥ 0. The two equal sides from O therefore force the other vertices to have the same x-coordinate; since they are distinct, they are (x, ±y), and each side from the origin makes 30° with the axis: y = x/√3. Then x2/3 = 4ax gives x = 12a, y = 12a/√3 = 4√3a, and the side is 2y = 8√3a.