Start learning →

Indian Navy Pilot Conic Sections — practice questions

Explore 32 free Conic Sections questions in Indian Navy Pilot Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

Practise Conic Sections with Quill →

Conic Sections questions and exercises with answers

Showing questions 1–30 of 32.

A parabola has its vertex at the origin, its axis along the positive $x$-axis, and passes through the point $(4, 6)$. Find its equation.Read the answer and worked solution →Common tangent to circle x² + y² = 5 and parabola y² = 8x:Read the answer and worked solution →Sum of distances from any point on ellipse 9x² + 16y² = 144 to its foci:Read the answer and worked solution →Two dice are rolled. The probability of getting AT LEAST ONE SIX is:Read the answer and worked solution →By Kepler's First Law, every planet orbits the Sun in:Read the answer and worked solution →Match the trig identity to its resultRead the answer and worked solution →Sort each inverse function into its domainRead the answer and worked solution →Tap every function whose domain is [−1, 1]Read the answer and worked solution →Match each inverse function to its domainRead the answer and worked solution →Match each inverse function to its principal value branchRead the answer and worked solution →Flip and pair: function with its principal rangeRead the answer and worked solution →Which principal ranges sit around zero, and which run 0 to π?Read the answer and worked solution →Tap every function whose principal range has a point removedRead the answer and worked solution →Name the principal value branchRead the answer and worked solution →Domain checkRead the answer and worked solution →Put the reasoning in order: why an inverse needs a restricted domainRead the answer and worked solution →True or false: existence of inversesRead the answer and worked solution →True or false: reading the tableRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Complete the summary tableRead the answer and worked solution →Complete the rule about restricting a domainRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Principal values worked in the chapterRead the answer and worked solution →Order the steps for finding a principal valueRead the answer and worked solution →True or false: principal valuesRead the answer and worked solution →Match each identity to the condition it needsRead the answer and worked solution →Assertion and reason: identitiesRead the answer and worked solution →Match each mathematician to the year your syllabus givesRead the answer and worked solution →

Worked practice questions

A parabola has its vertex at the origin, its axis along the positive $x$-axis, and passes through the point $(4, 6)$. Find its equation.

  1. $y^2 = 9x$
  2. $y^2 = 4x$
  3. $y^2 = 6x$
  4. $y^2 = 36x$
Show answer and solution

Correct answer: A. $y^2 = 9x$

**Setup.** Axis along positive $x$-axis with vertex at origin ⇒ standard form $y^2 = 4ax$. **Plug in the known point.** $(4, 6)$ on the curve: $6^2 = 4a \cdot 4 \Rightarrow 36 = 16a \Rightarrow a = \dfrac{9}{4}$. **Substitute back.** $y^2 = 4 \cdot \dfrac{9}{4} \cdot x = 9x$. Quick verification: $(4, 6)$ ⇒ $36 = 9 \times 4 = 36$ ✓. **Distractor check.** Option B ($y^2 = 4x$) corresponds to $a = 1$ — would pass through $(4, 4)$, not $(4, 6)$. Option D ($a = 9$) overshoots: $(4, ?) \Rightarrow y^2 = 144$, $y = 12$.

Open this question →

Common tangent to circle x² + y² = 5 and parabola y² = 8x:

  1. y = 0
  2. Two of form y = ±x + c via tangency conditions
  3. Cannot determine
  4. y = x + 2
Show answer and solution

Correct answer: B. Two of form y = ±x + c via tangency conditions

**Tangent to circle (radius sqrt(5)):** y = mx ± sqrt(5)sqrt(1+m²). Tangent to parabola y² = 8x (a=2): y = mx + 2/m. For common tangent, set 2/m = ±sqrt(5)sqrt(1+m²). **Squaring:** 4/m² = 5(1+m²) → 5m⁴ + 5m² - 4 = 0. Solutions for m² give specific common tangents.

Open this question →

Sum of distances from any point on ellipse 9x² + 16y² = 144 to its foci:

  1. Cannot determine
  2. 8 (= 2a, where a² = 16 → a = 4)
  3. 6
  4. 12
Show answer and solution

Correct answer: B. 8 (= 2a, where a² = 16 → a = 4)

x²/16 + y²/9 = 1. a² = 16, a = 4. Sum of focal distances = 2a = 8. (Definition of ellipse: 2a = constant.)

Open this question →

How to practise

  1. Choose a Conic Sections question and try to work out your answer.
  2. Open the question to check the correct answer and read its worked solution.
  3. Continue in the app for bite-sized lessons, explanations and review with Quill. You can start without creating an account.

More Mathematics topics

SequencesBinomial TheoremSequences and SeriesGeometryProbability & StatisticsVectors & 3D GeometryPermutations and CombinationsLinear Programming