Question

An ellipse has its centre at (1, −1) and semi-major axis = 8 and it passes through the point
(1, 3). The equation of the ellipse is

(a) $\frac{{\left(x+1\right)}^2}{64}+\frac{{\left(y+1\right)}^2}{16}=1$

(b) $\frac{{\left(x-1\right)}^2}{64}+\frac{{\left(y+1\right)}^2}{16}=1$

(c) $\frac{{\left(x-1\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{64}=1$

(d) $\frac{{\left(x+1\right)}^2}{64}+\frac{{\left(y-1\right)}^2}{16}=1$

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\frac{{\left(\mathrm{x}-1\right)}^2}{64}+\frac{{\left(\mathrm{y}+1\right)}^2}{16}=1 \\ \text{According to the question, the centre is at}\left(1,-1\right)\text{.} \\ a=8\left(\text{Given}\right) \\ \Rightarrow \frac{{\left(x-1\right)}^2}{a^2}+\frac{{\left(y+1\right)}^2}{b^2}=1...\left(1\right) \\ \text{It passes through point}\left(1,3\right)\text{.} \\ \mathrm{i}.\mathrm{e}.x=1\mathrm{and}y=3 \\ \mathrm{Putting}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get}: \\ \frac{{\left(1-1\right)}^2}{a^2}+\frac{{\left(3+1\right)}^2}{b^2}=1 \\ \Rightarrow \frac{16}{b^2}=1 \\ \Rightarrow b^2=16\mathrm{or}b=4 \\ \text{Substituting the values of}\mathit{\text{a}}\text{and}\mathit{\text{b}}\text{in eq. (1), we get:} \\ \frac{{\left(x-1\right)}^2}{64}+\frac{{\left(y+1\right)}^2}{16}=1 \end{gathered}$$