EllipseHardBloom L3
Question
If the focus, centre and eccentricity of an ellipse are respectively $(3,4)(2,3)$ and $\frac{1}{2}$, then its equation is given by:
Options
A.$\frac{(x + y - 5)^{2}}{16} + \frac{(x - y + 1)^{2}}{12} = 1$
B.$\frac{(x - y + 1)^{2}}{16} + \frac{(x + y - 5)^{2}}{12} = 1$
C.$\sqrt{(x - 3)^{2} + (y - 4)^{2}} + \sqrt{(x - 2)^{2} + (y - 1)^{2}} = 4\sqrt{2}$
D.$\sqrt{(x - 3)^{2} + (y - 4)^{2}} + \sqrt{(x - 1)^{2} + (y - 2)^{2}} = 4\sqrt{2}$
Solution
Sol. Equation of major axis is $y = x + 1$
Equation of minor axis is $x + y = 5$
$$\begin{matrix} ae & \ = \sqrt{(4 - 3)^{2} + (3 - 2)^{2}} \\ & \ = \sqrt{2} \Rightarrow a = 2\sqrt{2} \\ b^{2} & \ = a^{2} - a^{2}e^{2} = 8 - 2 = 6 \Rightarrow b = \sqrt{6} \end{matrix}$$
Equation of ellipse is
$$\frac{\left( \frac{x + y - 5}{\sqrt{2}} \right)^{2}}{(2\sqrt{2})^{2}} + \frac{\left( \frac{y - x - 1}{\sqrt{2}} \right)^{2}}{(\sqrt{6})^{2}} = 1$$
Also, its equation is
$$\sqrt{(x - 3)^{2} + (y - 4)^{2}} + \sqrt{(x - 1)^{2} + (y - 2)^{2}} = 2(2\sqrt{2})$$
Create a free account to view solution
View Solution FreeMore Ellipse Questions
If the line joining foci subtends an angle of $90^{\circ}$ at an extremity of minor axis then the eccentricity of the el...For the ellipse $5x^{2} + 9y^{2} + 10x - 36y - 4 = 0$ which of the following are true :...An ellipse has its center at $(1, - 2)$, one focus at $(3, - 2)$ and one vertex at (5,-2). Then the length of its latus ...If the latus rectum of an ellipse be equal to half of its minor axis, then its eccentricity is-...$x - 2y + 4 = 0$ is a common tangent to $y^{2} = 4x$ and $\frac{x^{2}}{4} + \frac{y^{2}}{b^{2}} = 1$. Then the value of ...