--- title: "Cone with vertex (3,1,2) and given base" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:cone-vertex-3-1-2" source: https://www.bctengineer.com/q/semester_2/mathematics/cone-vertex-3-1-2 site: BCT Engineering --- # Cone with vertex (3,1,2) and given base Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2075 Baisakh, 5 marks; 2069 Bhadra, 5 marks (asked 2 times) - Video: [Find the equation of cone whose vertex is (α,β,r) and guiding curve is ax^2+by^2=1, z=0.Top maths](https://www.youtube.com/watch?v=8i5bLufGoNI) — Doctor of Mathematics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/cone-vertex-3-1-2 - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Find the equation of cone with vertex at (3,1,2) and base 2x²+3y² = 1, z = 1. ## Short answer Lines through vertex (3,1,2) meet z=1 at (3−l/n, 1−m/n, 1). Enforcing 2x²+3y²=1 and using l/n=(x−3)/(z−2), m/n=(y−1)/(z−2) gives 2(3z−x−3)²+3(z−y−1)²=(z−2)². Expanding yields 2x²+3y²+20z²−12xz−6yz+12x+6y−38z+17=0. ## Answer **Cone.** A cone with vertex V(3, 1, 2) and base curve (guiding curve) 2x² + 3y² = 1, z = 1. #### Generators A line through the vertex with direction ratios (l, m, n): (x − 3)/l = (y − 1)/m = (z − 2)/n. ... (1) It meets the plane z = 1 where (z − 2)/n = −1/n. So the point of intersection is x = 3 − l/n, y = 1 − m/n, z = 1. This point must lie on the base curve: ```text 2(3 − l/n)² + 3(1 − m/n)² = 1 ... (2) ``` #### Eliminate l, m, n From (1), l/n = (x − 3)/(z − 2) and m/n = (y − 1)/(z − 2). Put in (2): ```text 2 [3 − (x − 3)/(z − 2)]² + 3 [1 − (y − 1)/(z − 2)]² = 1 2 [3(z − 2) − (x − 3)]² + 3 [(z − 2) − (y − 1)]² = (z − 2)² 2 (3z − x − 3)² + 3 (z − y − 1)² = (z − 2)² ``` #### Expand ```text 2(9z² + x² + 9 − 6xz − 18z + 6x) = 2x² − 12xz + 12x + 18z² − 36z + 18 3(z² + y² + 1 − 2yz − 2z + 2y) = 3y² − 6yz + 6y + 3z² − 6z + 3 (z − 2)² = z² − 4z + 4 ``` Sum of the first two minus the third: **2x² + 3y² + 20z² − 12xz − 6yz + 12x + 6y − 38z + 17 = 0** Check: at z = 1 this gives 2x² + 3y² − 1 = 0, the given base curve. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where does the generator through (3, 1, 2) meet z = 1? A. x = 3 - l/n, y = 1 - m/n B. x = 3 + l/n, y = 1 + m/n C. x = 3 - l n, y = 1 - m n D. x = 3 - 2l/n, y = 1 - 2m/n Answer: A. (z - 2)/n = -1/n, so x = 3 - l/n and y = 1 - m/n. (Key point: Point of intersection with plane) 2. What condition is applied to this point? A. It lies on the plane x = 3 B. It lies on the base curve 2x^2 + 3y^2 = 1 C. It lies on the axis of the cone D. It satisfies 2x^2 + 3y^2 = 2 Answer: B. The generator's intersection must be on the guiding curve. (Key point: Condition) 3. How are l, m, n eliminated? A. Replace l/n and m/n by x/z and y/z B. Replace l/n and m/n by (z - 2)/(x - 3) and (z - 2)/(y - 1) C. Replace l/n and m/n by (x - 3)/(z - 2) and (y - 1)/(z - 2) D. Replace l and m by x and y and n by 1 Answer: C. The generator's direction ratios are proportional to the coordinates of a point relative to the vertex. (Key point: Elimination) 4. What check shows the final equation is right? A. Putting z = 1 gives back 2x^2 + 3y^2 - 1 = 0 B. Putting z = 0 gives the base circle C. Putting x = 3 gives y = 1 D. Putting z = 2 gives the base curve Answer: A. The cone must cut z = 1 in the base curve. 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