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 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 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:
- Length of latus rectum of ellipse \[ \frac{x^2}{9} + \frac{y^2}{16} = 1 \]
- Find the center of the ellipse given by the equation \[ 4x^2 + 24x + 9y^2 - 18y + 9 = 0 \]
- If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is:
Questions Asked in CBSE CLASS XII exam
Rupal, Shanu and Trisha were partners in a firm sharing profits and losses in the ratio of 4:3:1. Their Balance Sheet as at 31st March, 2024 was as follows:
(i) Trisha's share of profit was entirely taken by Shanu.
(ii) Fixed assets were found to be undervalued by Rs 2,40,000.
(iii) Stock was revalued at Rs 2,00,000.
(iv) Goodwill of the firm was valued at Rs 8,00,000 on Trisha's retirement.
(v) The total capital of the new firm was fixed at Rs 16,00,000 which was adjusted according to the new profit sharing ratio of the partners. For this necessary cash was paid off or brought in by the partners as the case may be.
Prepare Revaluation Account and Partners' Capital Accounts.- CBSE CLASS XII - 2025
- Partnership Accounts
On the basis of the following hypothetical data, calculate the percentage change in Real Gross Domestic Product (GDP) in the year 2022 – 23, using 2020 – 21 as the base year.
Year Nominal GDP Nominal GDP (Adjusted to Base Year Price) 2020–21 3,000 5,000 2022–23 4,000 6,000 - CBSE CLASS XII - 2025
- National Income Accounting
- Determine those values of $x$ for which $f(x) = \frac{2}{x} - 5$, $x \ne 0$ is increasing or decreasing.
- CBSE CLASS XII - 2025
- Functions
- Analyse how freedom is curbed in the context of Franz in ‘The Last Lesson’ and Saheb in ‘Lost Spring’.
- CBSE CLASS XII - 2025
- The Last Lesson
- Altima Ltd. invited applications for 2,00,000 equity shares of ₹ 10 at a premium of ₹ 4 per share. Amount payable:
On application and allotment – ₹ 7 (incl. ₹ 1 premium)
On first and final call – Balance.
Applications received for 2,40,000 shares. 30,000 rejected. Manvi allotted 4,000 shares failed to pay first and final call. Her shares were forfeited. These were reissued at ₹ 4 per share fully paid-up.
Pass journal entries in the books of Altima Ltd.- CBSE CLASS XII - 2025
- Accounting for Share Capital
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}