Mathematics Class 11
🏠 Home Log in
TOPIC 33 OF 45

Hyperbola and Conic Sections Exercises

🎓 Class 11 Mathematics CBSE Theory Ch 10 — Conic Sections ⏱ ~30 min
🌐 Language: [gtranslate]

This MCQ module is based on: Hyperbola and Conic Sections Exercises

This mathematics assessment will be based on: Hyperbola and Conic Sections Exercises
Targeting Class 11 level in Coordinate Geometry, with Advanced difficulty.

Upload images, PDFs, or Word documents to include their content in assessment generation.

10.6 Hyperbola

Definition
A hyperbolai is the locus of points in a plane the absolute difference of whose distances from two fixed points (the foci) is a positive constant less than the distance between the foci.

Let the foci be \(F_1(-c,0)\) and \(F_2(c,0)\) and let the constant difference be \(2a\) with \(0<2a<2c\). For a point \(P\) on the curve, \(|PF_1-PF_2|=2a\). Define \(b^2=c^2-a^2\).

10.6.1 Derivation of standard equation

From \(\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=\pm 2a\), isolate and square twice. After simplification using \(b^2=c^2-a^2\):

Standard form (foci on x-axis)
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad c^2=a^2+b^2.\]

For foci on the \(y\)-axis: \(\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1.\)

(a,0) (−a,0) F₂(c,0) F₁(−c,0) y=(b/a)x y=−(b/a)x
Fig 10.7 — Hyperbola with two branches, foci, vertices and asymptotes \(y=\pm(b/a)x\).

10.6.2 Key quantities

  • Vertices: \((\pm a,0)\). Foci: \((\pm c,0)\).
  • Transverse axis along the \(x\)-axis, length \(2a\); conjugate axis along the \(y\)-axis, length \(2b\).
  • Eccentricity \(e=c/a>1\).
  • Latus rectum length \(=2b^2/a\).
  • Asymptotes (for \(\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1\)): \(y=\pm\tfrac{b}{a}x\). The two branches hug these straight lines at infinity.

10.6.3 Comparison of the three non-circular conics

FeatureParabolaEllipseHyperbola
Foci122
Eccentricity \(e\)\(=1\)\(0\(>1\)
Defining relation\(PF=PL\)\(PF_1+PF_2=2a\)\(|PF_1-PF_2|=2a\)
Shapeopen, 1 branchclosedopen, 2 branches
Example 14
Find all relevant quantities for the hyperbola \(\dfrac{x^2}{9}-\dfrac{y^2}{16}=1\).
\(a=3,b=4,c=\sqrt{9+16}=5\). Vertices \((\pm 3,0)\); foci \((\pm 5,0)\); \(e=\tfrac{5}{3}\); LR \(=\tfrac{32}{3}\); asymptotes \(y=\pm\tfrac{4}{3}x\).
Example 15
Find the equation of the hyperbola with foci \((0,\pm 12)\) and length of latus rectum 36.
Foci on \(y\)-axis: \(\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1\). \(c=12\). LR \(=\tfrac{2b^2}{a}=36\Rightarrow b^2=18a\). Also \(b^2=c^2-a^2=144-a^2\Rightarrow 18a=144-a^2\Rightarrow a^2+18a-144=0\Rightarrow a=\tfrac{-18+\sqrt{324+576}}{2}=\tfrac{-18+30}{2}=6\). So \(b^2=108\). Equation: \(\dfrac{y^2}{36}-\dfrac{x^2}{108}=1.\)
Example 16
Find the equation of the hyperbola with vertices \((\pm 2,0)\) and \(e=\tfrac{3}{2}\).
\(a=2,e=\tfrac{c}{a}\Rightarrow c=3\). \(b^2=c^2-a^2=9-4=5\). Equation: \(\dfrac{x^2}{4}-\dfrac{y^2}{5}=1.\)
Activity 10.4 — Hyperbolic positioning
Predict: Two listening posts detect a sound with a measurable time-difference of arrival. Which conic describes the set of all source locations consistent with this observation?
  1. Mark two listening posts at \((\pm 5,0)\) (km).
  2. Suppose the sound reaches the left post 4 km (of sound-travel) later than the right post — so \(|PF_1-PF_2|=4=2a\Rightarrow a=2\).
  3. \(c=5\Rightarrow b^2=25-4=21\). Draw the locus \(\tfrac{x^2}{4}-\tfrac{y^2}{21}=1\) — the source lies on one branch.
  4. Discuss how two more posts narrow down the source to a point (this is the LORAN navigation principle).
Insight: A constant difference of distances traces a hyperbola branch. Intersection of two such hyperbolas (from two pairs of stations) localizes the sound source — the geometric basis of LORAN.

In-text Exercises on the Hyperbola

Q1. Find all relevant quantities for \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\).
\(a=4,b=3,c=5\). Vertices \((\pm 4,0)\); foci \((\pm 5,0)\); \(e=\tfrac{5}{4}\); LR \(=\tfrac{9}{2}\); asymptotes \(y=\pm \tfrac{3}{4}x\).
Q2. Find the equation of the hyperbola with foci \((\pm 3\sqrt5,0)\) passing through \((\pm 10,0)\).
\(c=3\sqrt5\), vertices at \((\pm 10,0)\Rightarrow a=10,a^2=100\). \(b^2=c^2-a^2=45-100\) negative — impossible, so the given "foci" data conflicts. (Correcting: if vertices are \((\pm 10,0)\) we would need \(c>a\); this problem is typical of checking consistency.) If instead we read \(c^2=45\) and the curve passes through \((10,0)\) as a vertex, we'd require \(c>10\), violating data — no real hyperbola.
Q3. Find the equation of the hyperbola with vertices \((0,\pm 3)\) and \(e=\tfrac{5}{3}\).
\(a=3,c=ae=5\Rightarrow b^2=16\). Equation: \(\dfrac{y^2}{9}-\dfrac{x^2}{16}=1.\)
Q4. Find the equation of the hyperbola with foci \((\pm 5,0)\) and transverse axis length 8.
\(a=4,c=5,b^2=9\). Equation: \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1.\)

Miscellaneous End-of-Chapter Exercises

Q1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Set vertex at origin, axis along \(y\)-axis, opening up: \(x^2=4ay\). Rim: \((10,5)\): \(100=20a\Rightarrow a=5\). Focus \((0,5)\) — at the rim's plane, along the axis.
Q2. 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.
Centre at midpoint of span. \(a=4,b=2\), equation \(\tfrac{x^2}{16}+\tfrac{y^2}{4}=1\). Point 1.5 m from end = 2.5 m from centre: \(x=2.5\Rightarrow y^2=4(1-6.25/16)=4\cdot\tfrac{9.75}{16}=\tfrac{39}{16}\Rightarrow y\approx 1.56\) m.
Q3. A rod 12 cm long moves with its ends always touching the two perpendicular walls. Find the locus of the point on the rod that is 3 cm from the end in contact with the \(x\)-axis.
Let the rod make angle \(\theta\) with \(x\)-axis; lower end at \((12\cos\theta,0)\), upper at \((0,12\sin\theta)\). Point 3 cm from lower: \((9\cos\theta,3\sin\theta)\). Let \((x,y)=(9\cos\theta,3\sin\theta)\Rightarrow \tfrac{x^2}{81}+\tfrac{y^2}{9}=1\). An ellipse.
Q4. Find the equation of the ellipse whose centre is at origin, the major axis along the \(x\)-axis passing through \((4,3)\) and \((-1,4)\).
\(\tfrac{16}{a^2}+\tfrac{9}{b^2}=1\) and \(\tfrac{1}{a^2}+\tfrac{16}{b^2}=1\). Let \(u=1/a^2,v=1/b^2\): \(16u+9v=1,\ u+16v=1\). Solve: from first \(u=(1-9v)/16\); sub: \((1-9v)/16+16v=1\Rightarrow 1-9v+256v=16\Rightarrow 247v=15\Rightarrow v=15/247\Rightarrow u=(1-9\cdot 15/247)/16=(247-135)/(247\cdot 16)=112/3952=7/247\). So \(a^2=247/7,b^2=247/15\). Equation: \(\dfrac{7x^2+15y^2}{247}=1\), i.e. \(7x^2+15y^2=247.\)
Q5. A man running a racecourse notes that the sum of his distances from two flag-posts is always 10 m and the distance between the flag-posts is 8 m. Find the equation of the path traced.
Ellipse with \(2a=10,\ 2c=8\Rightarrow a=5,c=4,b^2=9\). Placing foci on \(x\)-axis centred at origin: \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1.\)
Q6. An arch is in the shape 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?
Vertex at origin, opens down: \(x^2=-4ay\). Base at \(y=-10,\ x=\pm 2.5\): \(6.25=40a\Rightarrow a=\tfrac{5}{32}\). At \(y=-2\): \(x^2=-4\cdot\tfrac{5}{32}\cdot(-2)=\tfrac{10}{8}=\tfrac{5}{4}\Rightarrow x=\tfrac{\sqrt5}{2}\). Width \(=\sqrt5\approx 2.24\) m.

Summary

  • Circle: \((x-h)^2+(y-k)^2=r^2\); locus of points equidistant from a centre.
  • Parabola: \(y^2=4ax\); focus \((a,0)\), directrix \(x=-a\), LR \(=4a\), \(e=1\).
  • Ellipse: \(\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1\) with \(b^2=a^2-c^2\); \(PF_1+PF_2=2a\); \(0
  • Hyperbola: \(\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1\) with \(c^2=a^2+b^2\); \(|PF_1-PF_2|=2a\); \(e>1\); asymptotes \(y=\pm(b/a)x\).
  • Eccentricity classifies all conics: circle (0), ellipse (01).
Competency-Based Questions — Hyperbolic Applications
A power-plant cooling tower is shaped like a hyperboloid of one sheet. A cross-section through the tower's axis yields a hyperbola centred at the origin with transverse axis along the \(x\)-axis. The waist (narrowest horizontal radius) measures 10 m, and at a vertical distance of 20 m above/below the waist, the radius is 15 m.
Q1. Take the cross-section in the \(xy\)-plane with \(x\) horizontal (radius) and \(y\) vertical. Set up \(\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1\) and find \(a\) and \(b\).
L3 Apply
Waist → \(a=10\). Using \((15,20)\): \(\tfrac{225}{100}-\tfrac{400}{b^2}=1\Rightarrow \tfrac{400}{b^2}=1.25\Rightarrow b^2=320\Rightarrow b\approx 17.89\) m.
Q2. Compute the tower's radius at a height of 40 m above the waist and compare with 15 m.
L4 Analyse
\(\tfrac{x^2}{100}-\tfrac{1600}{320}=1\Rightarrow \tfrac{x^2}{100}=6\Rightarrow x^2=600\Rightarrow x\approx 24.49\) m. Much larger than 15 m at 20 m height — the tower flares outward as we move away from the waist.
Q3. The designer claims that the hyperboloid offers greater structural stability than a cylinder of radius 15 m because the waist is narrower. Evaluate whether this claim holds from a geometric standpoint.
L5 Evaluate
Geometrically, the narrow waist concentrates the load-path near the axis and the generating straight lines (a hyperboloid of one sheet is doubly-ruled) provide tension along the surface. Though a cylinder has uniform radius, the hyperboloid combines reduced wind-profile at the waist with straight-line structural members — so the claim holds qualitatively; formal stability analysis would require stress calculations.
Q4. Design a second, slimmer tower cross-section with waist radius 8 m and radius 12 m at height 15 m. Write its equation in standard form.
L6 Create
\(a=8\Rightarrow a^2=64\). At \((12,15)\): \(\tfrac{144}{64}-\tfrac{225}{b^2}=1\Rightarrow \tfrac{225}{b^2}=\tfrac{80}{64}=1.25\Rightarrow b^2=180\). Equation: \(\dfrac{x^2}{64}-\dfrac{y^2}{180}=1.\)
Assertion–Reason Questions
Assertion (A): The eccentricity of every hyperbola is greater than 1.
Reason (R): For a hyperbola \(c>a\), so \(e=c/a>1\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. Defining inequality \(2a<2c\) gives \(c>a\Rightarrow e>1\).
Assertion (A): The asymptotes of \(\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1\) are \(y=\pm\tfrac{a}{b}x\).
Reason (R): As \(|x|\to\infty\), \(y\approx\pm\sqrt{(b^2/a^2)(x^2-a^2)}\to\pm(b/a)x\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. A is false — the correct asymptotes are \(y=\pm\tfrac{b}{a}x\) (not \(a/b\)). R correctly derives these.

Frequently Asked Questions

What is the standard equation of a hyperbola?
The standard equation is x^2/a^2 - y^2/b^2 = 1. The transverse axis (length 2 a) lies along the x-axis and the conjugate axis (length 2 b) along the y-axis.
Where are the foci of a hyperbola?
The foci of x^2/a^2 - y^2/b^2 = 1 are at (plus/minus c, 0) where c squared = a squared + b squared. Notice the plus sign, unlike the ellipse.
What is the eccentricity of a hyperbola?
For a hyperbola e = c/a and e is always greater than 1 (since c > a). A larger e means the hyperbola opens more widely.
What are the asymptotes of a hyperbola?
The asymptotes of x^2/a^2 - y^2/b^2 = 1 are the straight lines y = (b/a) x and y = -(b/a) x, which the branches approach at infinity.
What is the summary of Class 11 Chapter 10 Conic Sections?
Chapter 10 introduces the circle, parabola, ellipse and hyperbola as intersections of a cone with a plane and develops the standard equation, foci, axes and eccentricity for each curve.
How important is Chapter 10 for JEE?
Conic Sections is a very high-weightage JEE topic: typically 2-3 questions in JEE Main and multiple multi-part questions in JEE Advanced across parabola, ellipse and hyperbola.
AI Tutor
Mathematics Class 11
Ready
Hi! 👋 I'm Gaura, your AI Tutor for Hyperbola and Conic Sections Exercises. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.