Ellipse is a crucial topic in the JEE Mains syllabus under the chapter of conic sections. Here, we provide the last five years of JEE questions on ellipses along with their solutions in simple words, ensuring to use important keywords like JEE Ellipse Questions, Previous Year Questions on Ellipse, JEE Ellipse Solutions, and Ellipse PYQ.
Question: Find the length of the latus rectum of the ellipse given by the equation:
(x^2)/16 + (y^2)/9 = 1
Solution:
The general equation of an ellipse is:
(x^2)/(a^2) + (y^2)/(b^2) = 1
Here, a^2 = 16 and b^2 = 9, so a = 4 and b = 3.
For a horizontal ellipse, the length of the latus rectum is:
Latus Rectum = 2(b^2)/a
Substitute b^2 = 9 and a = 4:
Latus Rectum = 2(9)/4 = 18/4 = 4.5
Answer: The length of the latus rectum is 4.5 units.
Question:
The foci of an ellipse are at (-4, 0) and (4, 0). If the length of the major axis is 10, find the equation of the ellipse.
Solution:
The distance between the foci is 2c = 8, so c = 4.
The length of the major axis is 2a = 10, so a = 5.
For an ellipse, the relationship is:
c^2 = a^2 - b^2
Substitute c = 4 and a = 5:
4^2 = 5^2 - b^2 → 16 = 25 - b^2 → b^2 = 9
The equation of the ellipse is:
(x^2)/25 + (y^2)/9 = 1
Answer: The equation of the ellipse is:
(x^2)/25 + (y^2)/9 = 1
Question:
If the equation of an ellipse is (x^2)/49 + (y^2)/25 = 1, find its eccentricity.
Solution:
The formula for eccentricity e is:
e = sqrt(1 - (b^2)/(a^2))
Here, a^2 = 49 and b^2 = 25, so a = 7 and b = 5.
Substitute the values:
e = sqrt(1 - 25/49) = sqrt(24/49) = sqrt(24)/7
Simplify sqrt(24): sqrt(24) = 2sqrt(6)
So, e = 2sqrt(6)/7.
Answer: The eccentricity is 2sqrt(6)/7.
Question:
Find the equation of the ellipse whose vertices are at (0, ±6) and whose eccentricity is 1/2.
Solution:
For a vertical ellipse, the equation is:
(x^2)/(b^2) + (y^2)/(a^2) = 1
The length of the major axis is 2a = 12, so a = 6.
Eccentricity e = c/a → c = e × a = 1/2 × 6 = 3.
Use c^2 = a^2 - b^2:
3^2 = 6^2 - b^2 → 9 = 36 - b^2 → b^2 = 27
The equation of the ellipse is:
(x^2)/27 + (y^2)/36 = 1
Answer: The equation of the ellipse is:
(x^2)/27 + (y^2)/36 = 1
Question:
The ellipse (x^2)/16 + (y^2)/4 = 1 is rotated about the x-axis. Find the volume of the solid generated.
Solution:
When an ellipse is rotated about the x-axis, the volume is given by:
V = π × a × b^2
Here, a^2 = 16 and b^2 = 4, so a = 4 and b = 2.
V = π × 4 × 4 = 16π
Answer: The volume of the solid is 16π cubic units.
An ellipse is a geometrical shape in which each point is the same total distance from the two fixed points in a conic section. For JEE aspirants, the previous year's question with solutions on Ellipse is a valuable resource. Students will come across questions from prior JEE examinations while studying this part.
The concept of an ellipse is crucial for the JEE exam. The set of all points on a plane whose distances from two points add up to a constant is called an ellipse. We will come across definitions, ellipse equations, and properties in this article, which will aid students in gaining a thorough understanding of the issue.