Question

:)))))))))):)))):)))):)))))))))) solve it​

Answer 1

Answer:

Explanation:

Nothing Jo aapka ma'am kare woh likhde

Answer 2

Answer:

$\bold\red{(A){T}^{2}≥8}$

Explanation:

Equation of normal to the parabola,

${y}^{2}=4ax$

at the point $( a{t}^{2} ,2at )}$ is

$y + tx = 2at + a {t}^{3} \: \: \: \: \: \: .............(i)$

Now,

normal cuts the parabola again at $(a{T}^{2},2aT)$

Then,

$2aT + t\:a{T}^{2}=2at+a{t}^{3}$

$= > 2a(T - t) = - at( {T}^{2} - {t}^{2} ) \\ \\ = > 2 = - t(T + t) \\ \\ = > {t}^{2} + tT + 2 = 0$

Since,

t is real,

$= > {T}^{2} - 4 \times 2 \times 1 \geqslant 0 \\ \\ = > \bold{ {T}^{2} \geqslant 8}$

Hence,

Correct Option is (A)