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Parabola question

2022 · 29 Jul · Shift 1 · Q37
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Parabola question

2022 · 29 Jul · Shift 1 · Q37

JEE MainMathematicsParabolaMCQ+4 / −1
Let the focal chord of the parabola P:y2=4x\mathrm{P}: y^{2}=4 xP:y2=4x along the line L:y=mx+c,m>0\mathrm{L}: y=\mathrm{m} x+\mathrm{c}, \mathrm{m}\gt 0L: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\mathrm{H}: x^{2}-y^{2}=4H: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 :
  1. A
    262 \sqrt{6}26​
  2. B
    2142 \sqrt{14}214​
  3. C
    464 \sqrt{6}46​
  4. D
    4144 \sqrt{14}414​
View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS, 6\SQRT{2}

  1. Given objects
  • Parabola: P:y2=4xP: y^2=4xP:y2=4x
  • Line: L:y=mx+cL: y=mx+cL:y=mx+c, with m>0m>0m>0
  • LLL cuts the parabola at M,NM,NM,N and this chord is a focal chord of the parabola.
  • LLL is tangent to the hyperbola H:x2−y2=4H: x^2-y^2=4H:x2−y2=4.
  • OOO is the vertex of PPP, so O=(0,0)O=(0,0)O=(0,0).
  • Focus of PPP is (1,0)(1,0)(1,0), and since the chord is a focal chord, the line LLL passes through (1,0)(1,0)(1,0).
  • Focus of HHH on positive xxx-axis is F=(22,0)F=(2\sqrt2,0)F=(22​,0) because for x2−y2=4x^2-y^2=4x2−y2=4, we have a2=4,b2=4a^2=4, b^2=4a2=4,b2=4, hence c2=a2+b2=8c^2=a^2+b^2=8c2=a2+b2=8.
  1. Use focal chord condition for the parabola

Since LLL passes through the focus (1,0)(1,0)(1,0), 0=m(1)+c  ⟹  c=−m.0=m(1)+c \implies c=-m.0=m(1)+c⟹c=−m. So the line is y=mx−m=m(x−1).y=mx-m=m(x-1).y=mx−m=m(x−1).

  1. Use tangency condition with the hyperbola

For the hyperbola x2−y2=4,x^2-y^2=4,x2−y2=4, a line y=mx+cy=mx+cy=mx+c is tangent if c2=4(1−m2).c^2=4(1-m^2).c2=4(1−m2). This comes from substituting y=mx+cy=mx+cy=mx+c into the hyperbola and requiring discriminant =0=0=0.

Here c=−mc=-mc=−m, so m2=4(1−m2).m^2=4(1-m^2).m2=4(1−m2). Thus, 5m2=4  ⟹  m2=45  ⟹  m=255m^2=4 \implies m^2=\frac45 \implies m=\frac{2}{\sqrt5}5m2=4⟹m2=54​⟹m=5​2​ (since m>0m>0m>0).

Hence c=−25.c=-\frac{2}{\sqrt5}.c=−5​2​. So the line is y=25(x−1).y=\frac{2}{\sqrt5}(x-1).y=5​2​(x−1).

  1. Find points MMM and NNN of intersection with the parabola

Substitute y=mx−my=mx-my=mx−m into y2=4xy^2=4xy2=4x: [m(x−1)]2=4x.[m(x-1)]^2=4x.[m(x−1)]2=4x. Using m2=45m^2=\frac45m2=54​, 45(x−1)2=4x\frac45 (x-1)^2=4x54​(x−1)2=4x (x−1)2=5x (x-1)^2=5x(x−1)2=5x x2−2x+1=5xx^2-2x+1=5xx2−2x+1=5x x2−7x+1=0.x^2-7x+1=0.x2−7x+1=0. So the two xxx-coordinates are roots of this quadratic: x1,2=7±352.x_{1,2}=\frac{7\pm 3\sqrt5}{2}.x1,2​=27±35​​.

Now y=m(x−1)=25(x−1).y=m(x-1)=\frac{2}{\sqrt5}(x-1).y=m(x−1)=5​2​(x−1). For x1=7+352x_1=\frac{7+3\sqrt5}{2}x1​=27+35​​, x1−1=5+352,x_1-1=\frac{5+3\sqrt5}{2},x1​−1=25+35​​, y1=25⋅5+352=5+355=5+3.y_1=\frac{2}{\sqrt5}\cdot \frac{5+3\sqrt5}{2}=\frac{5+3\sqrt5}{\sqrt5}=\sqrt5+3.y1​=5​2​⋅25+35​​=5​5+35​​=5​+3. For x2=7−352x_2=\frac{7-3\sqrt5}{2}x2​=27−35​​, x2−1=5−352,x_2-1=\frac{5-3\sqrt5}{2},x2​−1=25−35​​, y2=25⋅5−352=5−3.y_2=\frac{2}{\sqrt5}\cdot \frac{5-3\sqrt5}{2}=\sqrt5-3.y2​=5​2​⋅25−35​​=5​−3. Thus the two points are

\qquad N\left(\frac{7-3\sqrt5}{2},\,\sqrt5-3\right).$$ Since $\sqrt5-3<0$, this is consistent with one point above and one below the $x$-axis. --- 5. **Area of quadrilateral $OMFN$** The vertices in order are $O(0,0), M, F(2\sqrt2,0), N$. Use the shoelace formula:

\text{Area}= \frac12\left| 0\cdot y_M + x_M\cdot 0 + 2\sqrt2\cdot y_N + x_N\cdot 0

\left(0\cdot x_M + y_M\cdot 2\sqrt2 + 0\cdot x_N + y_N\cdot 0\right) \right|.

So,So,So,

\text{Area}=\frac12\left|2\sqrt2(y_N-y_M)\right|=\sqrt2,|y_M-y_N|.

Now, $$y_M-y_N=(3+\sqrt5)-(\sqrt5-3)=6.$$ Hence $$\text{Area}=\sqrt2\cdot 6=6\sqrt2.$$ --- 6. **Match with options** $$6\sqrt2=\sqrt{72}=2\sqrt{18}$$ which is **not** equal to any of $$2\sqrt6,\;2\sqrt{14},\;4\sqrt6,\;4\sqrt{14}.$$ So the computed area is $$\boxed{6\sqrt2}.$$ --- 7. **Check by a cleaner geometric method** For parabola $y^2=4x$, a point is $(t^2,2t)$. A focal chord has endpoints corresponding to parameters $t$ and $-1/t$. Area of quadrilateral $OMFN$ with $F=(1,0)$ for parabola-focus would be easy, but here $F$ is focus of hyperbola, i.e. $(2\sqrt2,0)$. Using shoelace directly again gives $$\text{Area}=\frac12\cdot OF\cdot |y_M-y_N|=\frac12\cdot 2\sqrt2 \cdot 6=6\sqrt2,$$ since $MN$ intersects the $x$-axis at the parabola focus $(1,0)$ and the vertical separation of endpoints is $6$. This confirms the result. --- 8. **Conclusion** The correct area is $$\boxed{6\sqrt2},$$ which is not present in the options. Therefore the stored correct answer $B$ is inconsistent with the mathematics.
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