Q54 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics
The equation of the ellipse with its focus at \((6,2)\) centre at \((1,2)\) and which passes through the point \((4,6)\) is
- A\(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{16}=1\)
- B\(\frac{(x-1)^2}{25}+\frac{(y-2)^2}{20}=1\)
- C\(\frac{(x-1)^2}{45}+\frac{(y-1)^2}{16}=1\)
- Correct answerD\(\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1\)
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.
Hence, the correct answer is option: 'D'.