Q97 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics
If $\left(2 t^2, 4 t\right)$ is a point on the parabola $y^2=8 x$ such that its focal distance is 3 , then $t=$
- A$\pm 1$
- B$\pm \frac{1}{2}$
- C$\pm \frac{1}{\sqrt{3}}$
- Correct answerD$\pm \frac{1}{\sqrt{2}}$
Solution
Given equation of parabola $y^2=8 x$ with point $\left(2 \mathrm{t}^2, 4 \mathrm{t}\right)$
Foci coordinate $(\mathrm{a}, 0)=(2,0)$.
According to question,
$$
\begin{aligned}
& \sqrt{\left(2 \mathrm{t}^2-2\right)^2+(4 \mathrm{t}-0)^2}=3 \\
& 4 \mathrm{t}^4+4-8 \mathrm{t}^2+16 \mathrm{t}^2=9 \\
& 4 \mathrm{t}^4+8 \mathrm{t}^2-5=0 \\
& 4 \mathrm{t}^4+10 \mathrm{t}^2-2 \mathrm{t}^2-5=0 \\
& 2 \mathrm{t}^2\left(2 \mathrm{t}^2-1\right)\left(2 \mathrm{t}^2+5\right)=0 \\
& 2 \mathrm{t}^2-1=0, \quad 2 \mathrm{t}^2+5=0 \\
& \mathrm{t}^2=\frac{1}{2} \quad \mathrm{t}^2=-\frac{5}{2} \\
& \mathrm{t}= \pm \frac{1}{\sqrt{2}}
\end{aligned}
$$
So, option (d) is correct.
Hence, the correct answer is option: 'D'.