Q96 / 160 · TS EAMCET 2016 · Mathematics
For the ellipse given by $\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$, match the equations of the lines given in List I with those on the List II.
- A(a)
- B(b)
- C(c)
- Correct answerD(d)
Solution
Given equation of ellipse
$$
\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1
$$
Let $(x-3)=X$ and $(y-2)=Y$, then
$$
\frac{x^2}{5^2}+\frac{y^2}{4^2}=1
$$
Here, $\quad a=5$ and $b=4$.
Also, $\quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Now, equation of the major axis $Y=0$
$$
\Rightarrow
$$
$$
\begin{aligned}
y-2 & =0 \\
y & =2
\end{aligned}
$$
So, (i) $\rightarrow$ (q)
Equation of a directrix
$$
\begin{aligned}
& x & =\frac{a}{e}=\frac{5}{3} \times 5=\frac{25}{3} \\
\Rightarrow & x-3 & =\frac{25}{3} \\
\Rightarrow & x & =\frac{25}{3}+3=\frac{34}{3} \\
\Rightarrow & 3 x & =34
\end{aligned}
$$
So, (ii) $\rightarrow$ (p)
Equation of a latusrectum,
$$
\begin{aligned}
& x & = \pm \text { ae } \\
\Rightarrow & x-3 & = \pm \text { ae } \\
\Rightarrow & x-3 & = \pm 5 \times \frac{3}{5} \\
\Rightarrow & x & = \pm 3+3
\end{aligned}
$$
$$
\therefore \quad x=6,0
$$
So, (iii) $\rightarrow$ (s)
Hence, the correct answer is option: 'D'.