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Parabola question

2023 · Shift 1 · Q24
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Parabola question

2023 · Shift 1 · Q24

JEE AdvancedMathematicsParabolaMCQ+3 / −1
Let PPP be a point on the parabola y2=4axy^2=4 a xy2=4ax, where a>0a\gt 0a>0. The normal to the parabola at PPP meets the xxx-axis at a point QQQ. The area of the triangle PFQP F QPFQ, where FFF is the focus of the parabola, is 120 . If the slope mmm of the normal and aaa are both positive integers, then the pair (a,m)(a, m)(a,m) is
  1. A
    (2,3)(2,3)(2,3)
  2. B
    (1,3)(1,3)(1,3)
  3. C
    (2,4)(2,4)(2,4)
  4. D
    (3,4)(3,4)(3,4)
View written solutionFree

Correct answer: A

  1. Parametrize the point on the parabola

For the parabola y2=4ax,y^2=4ax,y2=4ax, a standard point is P(at2,2at).P(at^2,2at).P(at2,2at). The focus is F(a,0).F(a,0).F(a,0).

  1. Slope of the normal

For y2=4axy^2=4axy2=4ax, the tangent at parameter ttt is ty=x+at2,ty=x+at^2,ty=x+at2, so its slope is 1t.\frac{1}{t}.t1​. Hence the slope of the normal is −t.-t.−t. Given the slope of the normal is mmm, we have m=−t.m=-t.m=−t. Since m>0m>0m>0, this means t=−m.t=-m.t=−m.

Therefore the coordinates of PPP become P(am2,−2am).P(am^2,-2am).P(am2,−2am).

  1. Equation of the normal and point QQQ on the xxx-axis

The normal at parameter ttt to y2=4axy^2=4axy2=4ax is y=−tx+2at+at3.y=-tx+2at+at^3.y=−tx+2at+at3. Substitute t=−mt=-mt=−m: y=mx−2am−am3.y=mx-2am-am^3.y=mx−2am−am3.

To find where it meets the xxx-axis, put y=0y=0y=0: 0=mx−2am−am30=mx-2am-am^30=mx−2am−am3 mx=am(2+m2).mx=am(2+m^2).mx=am(2+m2). Since m>0m>0m>0, x=a(2+m2).x=a(2+m^2).x=a(2+m2). Thus, Q(a(2+m2),0).Q\big(a(2+m^2),0\big).Q(a(2+m2),0).

  1. Area of triangle PFQPFQPFQ

Points are: P(am2,−2am),F(a,0),Q(a(2+m2),0).P(am^2,-2am),\quad F(a,0),\quad Q(a(2+m^2),0).P(am2,−2am),F(a,0),Q(a(2+m2),0).

Since FFF and QQQ lie on the xxx-axis, the base FQFQFQ has length FQ=a(2+m2)−a=a(m2+1).FQ=a(2+m^2)-a=a(m^2+1).FQ=a(2+m2)−a=a(m2+1).

The perpendicular distance of PPP from the xxx-axis is ∣yP∣=2am.|y_P|=2am.∣yP​∣=2am.

So the area is 12⋅a(m2+1)⋅2am=a2m(m2+1).\frac12\cdot a(m^2+1)\cdot 2am= a^2m(m^2+1).21​⋅a(m2+1)⋅2am=a2m(m2+1). Given area =120=120=120, a2m(m2+1)=120.a^2m(m^2+1)=120.a2m(m2+1)=120.

  1. Check the options

  • A: (a,m)=(2,3)(a,m)=(2,3)(a,m)=(2,3) a2m(m2+1)=22⋅3⋅(9+1)=4⋅3⋅10=120.a^2m(m^2+1)=2^2\cdot 3\cdot (9+1)=4\cdot 3\cdot 10=120.a2m(m2+1)=22⋅3⋅(9+1)=4⋅3⋅10=120. Correct.

  • B: (1,3)(1,3)(1,3) 12⋅3⋅10=30≠120.1^2\cdot 3\cdot 10=30\neq 120.12⋅3⋅10=30=120. Incorrect.

  • C: (2,4)(2,4)(2,4) 22⋅4⋅17=272≠120.2^2\cdot 4\cdot 17=272\neq 120.22⋅4⋅17=272=120. Incorrect.

  • D: (3,4)(3,4)(3,4) 32⋅4⋅17=612≠120.3^2\cdot 4\cdot 17=612\neq 120.32⋅4⋅17=612=120. Incorrect.

  1. Conclusion

The required pair is (a,m)=(2,3).\boxed{(a,m)=(2,3)}.(a,m)=(2,3)​.

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