Let a tangent to the curve y ^ 2 = 24x meet the curve xy = 2 at the...

Given:

${y}^{2} = 24x$

Comparing with ${y}^{2} = 4ax$, we get

$4a = 24 ⇒ a = 6$

Also given $xy = 2$

Let any point $Q ≡ \left(a{t}^{2}, 2at\right) ≡$\left(6{t}^{2}, 12t\right)$on the parabola.</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/c291ff38-707a-453f-9003-b179de3df3f4-9893440.PNG" style="width: 362px; height: 242px;" /></p><p>Equation of tangent at$Q$is</p><p>$12yt = 12\left(x + 6{t}^{2}\right)$</p><p>$⇒ yt = x + 6{t}^{2}$</p><p>$⇒ x - yt + 6{t}^{2} = 0 . . . . \left(i\right)$</p><p>Let mid-point of$AB$be$P\left(h, k\right)$.</p><p>Equation of chord of hyperbola is</p><p>$\dfrac{xk + yh}{2} = hk$</p><p>$⇒ xk + hy - 2hk = 0 . . . \left(ii\right)$</p><p>Since, \left(i\right) \text{ and } \left(ii\right)$ are same, so on comparing, we get

$\dfrac{k}{1} = \dfrac{h}{- t} = \dfrac{- 2hk}{6{t}^{2}}$

$⇒ t = - \dfrac{h}{k} \text{ and }$$\dfrac{h}{- t} = \dfrac{- 2hk}{6{t}^{2}} ⇒ t = \dfrac{k}{3}$

So,

$\dfrac{k}{3} = - \dfrac{h}{k}$

^{$⇒ {k}^{2} = - 3h$}

Hence, locus is ${y}^{2} = - 3x$.

Therefore, directrix is $4x = 3$.

Length of latus rectum is $= 4a = 3$