JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let a tangent to the curve \(y^2=24 x\) meet the curve \(xy =2\) at the points \(A\) and \(B\). Then the mid points of such line segments \(A B\) lie on a parabola with the
- A directrix \(4 x=3\)
- B directrix \(4 x=-3\)
- C Length of latus rectum \(\frac{3}{2}\)
- D Length of latus rectum \(2\)
Answer & Solution
Correct Answer
(A) directrix \(4 x=3\)
Step-by-step Solution
Detailed explanation
\(y^2=24 x\) \(a=6\) \(x y=2\) \(AB \equiv ty = x +6 t ^2\) \(AB \equiv T = S _1\) \(kx + hy =2 hk \ldots\) From \((1)\) and \((2)\) \(\frac{ k }{1}=\frac{ h }{- t }=\frac{2 hk }{-6 t ^2}\) \(\Rightarrow\) then locus is \(y^2=-3 x\) Therefore directrix is \(4 x=3\)
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