Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse \(16x^2 + y^2 = 16\).
Solution and Explanation
The given equation is:
\(16x^2 + y^2 = 16\)
or \(\frac {x^2}{1}+ \frac {y^2}{16} = 1\)
or \(\frac {x^2}{12} + \frac {y^2}{42} = 1\) .......(1)
Here, the denominator of \(\frac {y^2}{4^2}\) is greater than the denominator of \(\frac {x^2}{1^2}\).
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation (1) with \(\frac {x^2}{b^2} + \frac {y^2}{a^2} = 1\), we obtain b = 1 and a = 4.
∴ \(c = \sqrt {(a^2 – b^2)}\)
\(c= \sqrt {(16-1)}\)
\(c= \sqrt {15}\)
Therefore,
The coordinates of the foci are \((0, ±\sqrt {15})\).
The coordinates of the vertices are (0, ±4).
Length of major axis = 2a = 8
Length of the minor axis = 2b = 2
Eccentricity, \(e = \frac ca = \frac {\sqrt {15}}{4}\)
Length of latus rectum = \(\frac {2b^2}{a} = \frac {(2×1^2)}{4} = \frac 24 = \frac 12\)
Top Questions on Ellipse
- Let $ e_1 $ and $ e_2 $ be the eccentricities of the ellipse $ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $ and the hyperbola $ \frac{x^2}{16} - \frac{y^2}{b^2} = 1, $ respectively. If $ b<5 $ and $ e_1 e_2 = 1 $, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:
- If the equation of an ellipse \( E \) is \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \), then the length of the latus rectum of \( E \) is?
- Let the product of the focal distances of the point $$ \left( \sqrt{3}, \frac{1}{2} \right) $$ on the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a > b), $$ be $ \frac{7}{4} $. Then the absolute difference of the eccentricities of two such ellipses is:
- Let the ellipse \( E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), \( a > b \) and \( E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \), \( A < B \), have the same eccentricity \( \sqrt{\frac{1}{3}} \). Let the product of their lengths of latus rectums be \( \sqrt{\frac{32}{3}} \) and the distance between the foci of \( E_1 \) be 4. If \( E_1 \) and \( E_2 \) meet at \( A \), \( B \), \( C \), and \( D \), then the area of the quadrilateral ABCD equals:
- Let the length of a latus rectum of an ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ be 10. If its eccentricity is $ e $, and the minimum value of the function $ f(t) = t^2 + t + \frac{11}{12} $, where $ t \in \mathbb{R} $, then $ a^2 + b^2 $ is equal to:
Questions Asked in CBSE CLASS XII exam
- You are provided with a large number of 1 \(\mu\text{F}\) identical capacitors and a power supply of 1200 V. The dielectric medium used in each capacitor can withstand up to 200 V only. Find the minimum number of capacitors and their arrangement required to build a capacitor system of equivalent capacitance of 2 \(\mu\text{F}\) for use with this supply.
- CBSE CLASS XII - 2025
- Combination of capacitors
- A parallel plate capacitor with plate area \( A \) and plate separation \( d \) has a capacitance \( C_0 \). A slab of dielectric constant \( K \) having area \( A \) and thickness \( \left(\frac{d}{4}\right) \) is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.
- CBSE CLASS XII - 2025
- capacitor with a dielectric
- An electric dipole of dipole moment \( 1.0 \times 10^{-12} \ \text{Cm} \) lies along the x-axis. An electric field of magnitude \( 2.0 \times 10^4 \ \text{NC}^{-1} \) is switched on at an instant in the region. The unit vector along the electric field is \( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \). The magnitude of the torque acting on the dipole at that instant is:
- The maximum value of \(Z = 3x + 4y\) subject to the constraints \(x + y \leq 1\), \(x \geq 0\), \(y \geq 0\) is:
- CBSE CLASS XII - 2025
- CBSE Compartment XII - 2025
- Linear Programming
Observe the given sequence of nitrogenous bases on a DNA fragment and answer the following questions:
(a) Name the restriction enzyme which can recognise the DNA sequence.
(b) Write the sequence after restriction enzyme cut the palindrome.
(c) Why are the ends generated after digestion called as ‘Sticky Ends’?- CBSE CLASS XII - 2025
- Biotechnology
Concepts Used:
Ellipse
Ellipse Shape
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Properties
- Ellipse has two focal points, also called foci.
- The fixed distance is called a directrix.
- The eccentricity of the ellipse lies between 0 to 1. 0≤e<1
- The total sum of each distance from the locus of an ellipse to the two focal points is constant
- Ellipse has one major axis and one minor axis and a center
Read More: Conic Section
Eccentricity of the Ellipse
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
Area of an ellipse
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Position of point related to Ellipse
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}