Miscellaneous Exercise — Conic Sections

1. M.13 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

   Hint. Place the vertex at the origin with the axis along the positive x-axis. At the depth x=5, the diameter is 20, so y=+-10 there.

   With vertex at the origin and axis along the x-axis, the equation is y^2=4ax. At x=5 (the depth), the diameter is 20, so y=10 (half the diameter). Substituting: 100=4a(5)=20a, giving a=5.

   ✦ Working through each part gives: The focus is at (5, 0), i.e. 5 cm from the vertex along the axis.

2. M.24 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

   Hint. Place the vertex at the top of the arch with the axis pointing down (positive y measured downward). At the base, y=10 and the half-width is 2.5.

   With the vertex at the top and y measured downward, the equation is x^2=4ay. At the base, y=10 and half-width x=2.5, so (2.5)^2=4a(10), giving 6.25=40a, i.e. a=5/32. At y=2 (2 m from the vertex): x^2=4a(2)=8(5/32)=5/4, so x=sqrt(5)/2. The full width there is 2x.

   ✦ Working through each part gives: Width = sqrt(5) metres, approximately 2.24 m.

3. M.35 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway, which is horizontal and 100 m long, is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

   Hint. Place the vertex of the parabola at the lowest point of the cable (above the middle of the roadway), with x measured horizontally from the middle. The cable's rise above the vertex at the ends (x=50) equals the difference between the longest and shortest wires.

   With vertex at the lowest point, the equation is x^2=4ay, where y is the cable's rise above the vertex. At x=50 (the end of the 100 m roadway), the rise equals 30-6=24 m, so 50^2=4a(24), giving 2500=96a, i.e. a=625/24. At x=18: y=x^2/(4a)=18^2/(4(625/24))=324/(625/6)=324(6)/625=1944/625. The wire length there is the shortest wire (6 m) plus this rise.

   ✦ Working through each part gives: Wire length = 6 + 1944/625 = 5694/625 metres, approximately 9.11 m.

4. M.44 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

   Hint. The full ellipse has semi-major axis a=4 (half the 8 m width) and semi-minor axis b=2 (the height at the centre). A point 1.5 m from one end (x=4) is at x=4-1.5=2.5.

   The ellipse is x^2/16+y^2/4=1, with a=4, b=2. At x=2.5: y^2=4(1-(2.5)^2/16)=4(1-6.25/16)=4(39/64)=39/16, so y=sqrt(39)/4.

   ✦ Working through each part gives: Height = sqrt(39)/4 metres, approximately 1.56 m.

5. M.54 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

   Hint. Let A be the end on the x-axis and B the end on the y-axis, with AB=12. If AP=3, then PB=9. Using the same right-triangle method as the textbook's own rod-locus example, the coordinates of P give cos(theta)=x/PB and sin(theta)=y/AP.

   Let A be on the x-axis, B on the y-axis, and P(x,y) on the rod with AP=3, so PB=12-3=9. Dropping perpendiculars from P onto both axes and using the right triangles formed with the rod: cos(theta)=x/9 and sin(theta)=y/3. Since cos^2(theta)+sin^2(theta)=1: (x/9)^2+(y/3)^2=1.

   ✦ Working through each part gives: x^2/81+y^2/9=1.

6. M.63 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the area of the triangle formed by the lines joining the vertex of the parabola x^2=12y to the ends of its latus rectum.

   Hint. Find a from the equation, locate the latus rectum's endpoints (at y=a), and apply the coordinate area formula with the vertex.

   Comparing x^2=12y with x^2=4ay gives a=3. The latus rectum is the line y=3, with endpoints where x^2=12(3)=36, i.e. x=+-6, so the endpoints are (6,3) and (-6,3). Using the vertex (0,0) and these two points in the area formula: Area=(1/2)|0(3-3)+6(3-0)+(-6)(0-3)|=(1/2)|0+18+18|=(1/2)(36)=18.

   ✦ Working through each part gives: Area = 18 square units.

7. M.74 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

   Hint. A constant sum of distances from two fixed points is exactly the definition of an ellipse, with the flag posts as the foci.

   The sum of distances is 2a=10, so a=5. The distance between the foci is 2c=8, so c=4. Then b^2=a^2-c^2=25-16=9. Placing the flag posts symmetrically on the x-axis gives the standard form.

   ✦ Working through each part gives: x^2/25+y^2/9=1.

8. M.85 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   An equilateral triangle is inscribed in the parabola y^2=4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

   Hint. By symmetry, the other two vertices are (x1,y1) and (x1,-y1) on the parabola. For an equilateral triangle, the distance from the origin to (x1,y1) must equal the distance between (x1,y1) and (x1,-y1), which is 2y1.

   Let the other two vertices be P(x1,y1) and Q(x1,-y1), so PQ=2y1. For an equilateral triangle, OP=PQ: OP^2=x1^2+y1^2=(2y1)^2=4y1^2, giving x1^2=3y1^2. Since P lies on the parabola, y1^2=4a.x1. Substituting: x1^2=3(4a.x1)=12a.x1, so x1=12a (dividing by x1, since x1 is not zero). Then y1^2=4a(12a)=48a^2, so y1=4a.sqrt(3). The side length is PQ=2y1.

   ✦ Working through each part gives: Side length = 8a.sqrt(3).

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh110.pdf) — Exercise 10.1 (15 questions), Exercise 10.2 (12 questions), Exercise 10.3 (20 questions), Exercise 10.4 (15 questions), plus the chapter's Miscellaneous Exercise (8 questions), 70 questions total. Confirmed against the CBSE curriculum PDF that Conic Sections has no formative-only carve-out at all, unlike its neighbouring chapters Straight Lines and Introduction to Three-Dimensional Geometry — every phrase in its syllabus line is summative. Confirmed the book teaches the general circle equation only through completing the square (Example 3), never naming g, f, c as constants, and never mentions hyperbola asymptotes anywhere (zero hits for the word itself, and none of Exercise 10.4's 15 questions ask for one). Exercise 10.1 Q3's centre coordinates were cross-checked via a 300dpi page render after the raw PDF text extraction reversed their order. Several exercise answers (10.1 Q10/Q11, 10.4 Q15, Misc Q2/Q3/Q5/Q8) were cross-verified against known standard results for this exact problem set.. Questions are referenced from the NCERT textbook for identification.