The equation of the ellipse with its focus at \((6,2)\) centre at \((1,2)\) and which passes through the…
The equation of the ellipse with its focus at \((6,2)\) centre at \((1,2)\) and which passes through the point \((4,6)\) is
- \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{16}=1\)
- \(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{20}=1\)
- \(\frac{(x-1)^2}{45}+\frac{(y-1)^2}{16}=1\)
- \(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
Solution
Given,
Focus \(\mathrm{S}=(6,2)\)
Centre \(\mathrm{C}=(1,2)=(h, k)\) say
Point \(\mathrm{P}=(4,6)\)
Required Equation of ellipse is
\(\frac{(x-1)^2}{a^2}+\frac{(y-2)^2}{b^2}=1\) ...(i)
Since, Eq. (i) passes through \(\mathrm{P}(4,6)\)
\(\begin{aligned}
\frac{(4-1)^2}{a^2}+\frac{(6-2)^2}{b^2} & =1 \\
\frac{9}{a^2}+\frac{16}{b^2} & =1 \quad \ldots (ii)
\end{aligned}\)
Since, Focus \(=(6,2)\)
\(\begin{aligned}
(h+a e, k) & =(6,2) \\
\therefore \quad h+a e & =6, \\
k & =2 \\
1+a e & =6 \\
a e & =5 \\
a^2 e^2 & =25 \\
b^2 & =a^2\left(1-e^2\right) \\
b^2 & =a^2-a^2 e^2 \\
b^2 & =a^2-25 \\
a^2 & =b^2+25 \quad \ldots (iii)
\end{aligned}\)
Put \(a^2\) value in Eq. (ii),
\(\begin{aligned}
\frac{9}{b^2+25}+\frac{16}{b^2} & =1 \\
9 b^2+16\left(b^2+25\right) & =b^2\left(b^2+25\right) \\
9 b^2+16 b^2+400 & =b^4+25 b^2
\end{aligned}\)
\(\begin{aligned}
b^4 & =400 \\
b^2 & =20
\end{aligned}\)
From Eq. (iii),
\(\begin{aligned}
& a^2=20+25 \\
& a^2=45
\end{aligned}\)
Put \(a^2, b^2\) Value in Eq. (i),
\(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
\([\therefore\) Answer written in the paper was wrong in RHS it should be 1]
Hence, option (d) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
Practice more Conic Sections questions on Aicharya