Let the focal chord of the parabola P:y2=4x along the line L:y=mx+c,m>0 meet the parabola at the points M and N . Let the line L be a tangent to the hyperbola H:x2−y2=4. If O is the vertex of P and F is the focus of H on the positive x -axis, then the area of the quadrilateral OMFN is :
(A)26
(B)214✓ Correct
(C)46
(D)414
Answer: (B)
Solution
The parabola y2=4x has focus at (1,0). Since L is a focal chord, it passes through (1,0), so c=−m.
The hyperbola 4x2−4y2=1 has focus at F(22,0). For line y=mx+c to be tangent to this hyperbola, c2=4m2−4. Substituting c=−m: m2=4m2−4, giving m=32 (since m>0).
The line is y=32x−32. Substituting into y2=4x: 34(x−1)2=4x, which gives x2−5x+1=0. The corresponding y-values satisfy y2−23y−4=0.
The area of quadrilateral OMFN with O at origin, F at (22,0), and M, N having y-coordinates y1,y2 is:
21∣22(y2−y1)∣=2(y2−y1)2=212+16=214
JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option