Previous year questions: Conic Sections

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Six items are real JEE Main 2026 questions from the official National Testing Agency papers and four are from the NCERT Exemplar for Class 11 Mathematics. Each heading shows the source and the marks. Stems were checked against the official PDF pages, and every numerical answer was re-derived and matched to the official key. Write the standard form of the conic before using any data from the question.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

An ellipse x squared by a squared plus y squared by b squared equal to 1, with a less than b, passes through (4, 3) and has eccentricity root 5 by 3. Find the length of its latus rectum.

Answer: Because b exceeds a here, the longer axis lies along y. The eccentricity relation then reads e squared equal to 1 minus a squared by b squared. With e equal to root 5 by 3, a squared by b squared is 4 by 9, so write a squared as 4k and b squared as 9k. The point (4, 3) gives 16 by 4k plus 9 by 9k equal to 1, that is 4 by k plus 1 by k equal to 1, so k is 5. Then a squared is 20 and b squared is 45. The latus rectum is 2a squared by b, which is 40 by 3 root 5, that is 8 root 5 by 3. The official key is 8 root 5 by 3.

Q2 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

Let P(3 cos alpha, 2 sin alpha), with alpha not zero, be a point on the ellipse x squared by 9 plus y squared by 4 equal to 1. Let Q be a point on the circle x squared plus y squared minus 14x minus 14y plus 82 equal to 0 and R a point on the line x plus y equal to 5 such that the centroid of triangle PQR is (2 plus cos alpha, 3 plus 2 by 3 sin alpha). Find the sum of the ordinates of all possible points R.

Answer: The circle has centre (7, 7). The centroid condition gives Q plus R equal to three times the centroid minus P. The ordinate equation is yQ plus yR equal to 9 plus 2 sin alpha minus 2 sin alpha, so yQ plus yR is 9 and yR is 9 minus yQ. Writing Q as (xQ, yQ) on the circle and R as (xR, yR) on the line, the circle condition becomes a quadratic in the two coordinates, and the line condition fixes the other. Solving gives yQ values whose corresponding yR values add to 8. The official key is 8.

Q3 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

Let H be the hyperbola x squared by a squared minus y squared by b squared equal to 1 with distance between its foci equal to 6 and distance between its directrices equal to 8 by 3. The line x equal to alpha meets H at A and B such that the area of triangle AOB is 4 root 15, where O is the origin. Find alpha squared.

Answer: The foci distance is 2c equal to 6, so c is 3. The directrices are x equal to plus or minus a by e, so their distance is 2a by e equal to 2a squared by c, which is 8 by 3. With c equal to 3 this gives a squared equal to 4, so a is 2 and b squared is c squared minus a squared equal to 5. The vertical line x equal to alpha cuts the hyperbola where y squared equals b squared times (alpha squared by a squared minus 1). The area of triangle AOB is half the base AB times the distance from O to the line, so 4 root 15 equals alpha times the absolute y value, and squaring with a squared equal to 4 and b squared equal to 5 gives alpha squared equal to 16. The official key is 16.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

The directrix of the parabola P: y squared equal to 8x cuts the x-axis at A. Let B(alpha, beta), with beta greater than 1, be a point on P such that the slope of AB is 3 by 5. If BC is a focal chord of P, find six times the area of triangle ABC.

Answer: The focus is (2, 0) and the directrix is x equal to minus 2, so A is (minus 2, 0). The slope condition gives beta equals 3 by 5 times (alpha plus 2). Substituting in beta squared equal to 8alpha gives 9(alpha plus 2) squared equal to 200 alpha, which solves to alpha equal to 18 (the other root is below 1). Then beta is 12. In the parameter form (2t squared, 4t), B has t equal to 3, and the other end of the focal chord has parameter minus 1 by 3, which is C(2 by 9, minus 4 by 3). The area of ABC is 80 by 3, so six times it is 160. The official key is 160.

Q5 (JEE Main 2026, 6 Apr Shift 1, 5 marks)

Let the centre of the circle x squared plus y squared plus 2gx plus 2fy plus 25 equal to 0 lie in the first quadrant on the line 2x minus y equal to 4. Let the area of an equilateral triangle inscribed in the circle be 27 root 3. Find the square of the length of the chord of the circle on the line x equal to 1.

Answer: The radius r satisfies (3 root 3 by 4) r squared equal to 27 root 3, so r squared is 36 and r is 6. The centre (minus g, minus f) is (a, 2a minus 4). Using the circle equation, the radius squared equals a squared plus (2a minus 4) squared minus 25, so a squared plus (2a minus 4) squared equal to 61. Expanding gives 5a squared minus 16a minus 45 equal to 0, whose positive root is a equal to 5. The centre is (5, 6). The chord on x equal to 1 is at distance 4 from the centre, so half the chord is the square root of 36 minus 16, that is 2 root 5, and the full chord is 4 root 5. Its square is 80. The official key is 80.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Let C be a circle with centre in the first quadrant touching the x-axis at a distance of 3 units from the origin. If C has an intercept of length 6 root 3 on the y-axis, find the length of the chord of C on the line x minus y equal to 3.

Answer: The centre is (3, r) with radius r. A chord of length 6 root 3 on the y-axis is at distance 3 from the centre, so half the chord satisfies r squared minus 9 equal to 27, giving r squared equal to 36 and r equal to 6. The centre is (3, 6). The line x minus y equal to 3 has distance from (3, 6) equal to the absolute value of 3 minus 6 minus 3 over root 2, that is 3 root 2. Half the required chord is the square root of 36 minus 18, that is 3 root 2, so the chord is 6 root 2. The official key is 6 root 2.

Q7 (NCERT Exemplar, 3 marks)

Find the coordinates of the centre of the circle passing through (0, 0), (a, 0) and (0, b).

Answer: The triangle formed by the three points is right-angled at the origin, so the centre is the midpoint of the hypotenuse joining (a, 0) and (0, b). The centre is (a by 2, b by 2). Source: NCERT Exemplar Class 11 Mathematics, Chapter 11, Exercise 11.3, Short Answer Type, Q3.

Q8 (NCERT Exemplar, 3 marks)

If one end of a diameter of the circle x squared plus y squared minus 4x minus 6y plus 11 equal to 0 is (3, 4), find the other end.

Answer: Completing the squares gives (x minus 2) squared plus (y minus 3) squared equal to 2, so the centre is (2, 3). If one end of the diameter is (3, 4), the other end is such that the centre is the midpoint: the other end is (2 times 2 minus 3, 2 times 3 minus 4), that is (1, 2). Source: NCERT Exemplar Class 11 Mathematics, Chapter 11, Exercise 11.3, Short Answer Type, Q7.

Q9 (NCERT Exemplar, 3 marks)

Given the ellipse 9x squared plus 25y squared equal to 225, find its eccentricity and foci.

Answer: Dividing by 225 gives x squared by 25 plus y squared by 9 equal to 1, so a is 5 and b is 3. Then c squared is 25 minus 9 equal to 16, c is 4 and the eccentricity is 4 by 5. The foci are (plus or minus 4, 0). Source: NCERT Exemplar Class 11 Mathematics, Chapter 11, Exercise 11.3, Short Answer Type, Q12.

Q10 (NCERT Exemplar, 3 marks)

If the line y equal to mx plus 1 is tangent to the parabola y squared equal to 4x, find the value of m.

Answer: The tangency condition for y equal to mx plus c on y squared equal to 4ax is c equal to a by m. Here c is 1 and a is 1, so 1 equals 1 by m, giving m equal to 1. Source: NCERT Exemplar Class 11 Mathematics, Chapter 11, Exercise 11.3, Short Answer Type, Q19.