Question

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. \(2\sqrt6\)
• B. \(2\sqrt{14}\)
• C. \(4\sqrt6\)
• D. \(4\sqrt{14}\)

Answer 1

The Correct Option is B

\(\begin{array}{l} H:\frac{x^2}{4}-\frac{y^2}{4}=1\end{array}\), Focus (ae, 0)
\(\begin{array}{l} F\left(2\sqrt{2},0\right) \end{array}\)

y = mx + c passes through (1, 0)
0 = m + c …(i)
L is tangent to hyperbola
\(\begin{array}{l} c=\pm\sqrt{4m^2-4}\end{array}\)
\(\begin{array}{l} -m=\pm\sqrt{4m^2-4}\end{array}\)
m² = 4m² – 4
\(\begin{array}{l} m=\frac{2}{\sqrt{3}}\end{array}\)
\(\begin{array}{l} c=\frac{-2}{\sqrt{3}} \end{array}\)
\(\begin{array}{l} T:y=\frac{2}{\sqrt{3}}x-\frac{2}{\sqrt{3}} \end{array}\)
P : y² = 4x
\(\begin{array}{l} y^2=4\left(\frac{\sqrt{3}y+2}{2}\right)\end{array}\)
\(\begin{array}{l} y^2-2\sqrt{3}y-4 =0\end{array}\)
\(\begin{array}{l}\text{Area}= \frac{1}{2}\begin{vmatrix}0 & 0 \\x_1 & y_1 \\2\sqrt{2} & 0 \\x_2 & y_2 \\0 & 0 \\\end{vmatrix}\end{array}\)
\(\begin{array}{l} =\left|\frac{1}{2}\left(-2\sqrt{2}y_1+2\sqrt{2}y_2\right)\right|\end{array}\)
\(\begin{array}{l} =\sqrt{2}\left|y_2-y_1\right|=\sqrt{2}\sqrt{\left(y_1+y_2\right)^2-4y_1y_2}\end{array}\)
\(\begin{array}{l} =\sqrt{56}\end{array}\)
\(\begin{array}{l} =2\sqrt{14}\end{array}\)