Ellipse
Properties of Ellipse: Axes, Latus Rectum, and Foci
GRB_1000_MCQ
Grade Class 12
Question:
The ends of the major axis of ellipse are $(-2, 4)$ and $(2, 1)$. If the point $(1, 3)$ lies on the ellipse. Then:
The length of major axis is equal to 10.
The length of minor axis is equal to $\dfrac{10}{\sqrt{24}}$.
The length of latus rectum of ellipse is $\dfrac{5}{6}$.
Square of the distance between the focii of ellipse is $\dfrac{125}{6}$.
Step-by-Step Solution
Key Concept: To determine the properties of an ellipse from its major axis endpoints and a point lying on it, first find the center and the length of the semi-major axis ($a$) using the distance formula and midpoint formula. Then, transform the coordinates of the given point to the principal axis system of the ellipse and use the standard equation $\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1$ to calculate the semi-minor axis length ($b$). All other ellipse properties can then be derived from $a$ and $b$.
Step 1: Determine the center and semi-major axis length.
The center of the ellipse is the midpoint of the given major axis ends $(-2, 4)$ and $(2, 1)$.
$$ \text{Center} = \left(\frac{-2+2}{2}, \frac{4+1}{2}\right) = \left(0, \frac{5}{2}\right) $$
The length of the major axis is $2a=10$.
Thus, the semi-major axis length is $a=5$.
Step 2: Determine the semi-minor axis length.
The length of the minor axis is $2b=\dfrac{10}{\sqrt{24}}$.
Thus, the semi-minor axis length is $b=\dfrac{5}{\sqrt{24}}$, which implies $b^2=\dfrac{25}{24}$.
Step 3: Determine the length of the latus rectum.
The length of the latus rectum of the ellipse is $\dfrac{5}{6}$.
Step 4: Determine the square of the distance between the foci.
The square of the distance between the foci is $(2c)^2 = \dfrac{125}{6}$.
This implies $4c^2 = \dfrac{125}{6}$, so $c^2 = \dfrac{125}{24}$.
Correct Answer: 1, 2, 3, 4