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Parabola question

2017 · Shift 1 · Q25
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  5. /2017 · Shift 1 · Q25

Parabola question

2017 · Shift 1 · Q25

JEE AdvancedMathematicsParabolaMultiple correct+4 / −1
If a chord, which is not a tangent, of the parabola y2 = 16x has the equation 2x + y = p, and mid-point (h, k), then which of the following is(are) possible value(s) of p, h and k?
  1. A
    p = −-− 1, h = 1, k = −-− 3
  2. B
    p = 2, h = 3, k = −-− 4
  3. C
    p = −-− 2, h = 2, k = −-− 4
  4. D
    p = 5, h = 4, k = −-− 3
View written solutionFree

Correct answer: B

  1. Given parabola and chord

    The parabola is y2=16xy^2=16xy2=16x which is of the form y2=4axy^2=4axy2=4ax with 4a=16  ⟹  a=4.4a=16 \implies a=4.4a=16⟹a=4.

    The chord has equation 2x+y=p.2x+y=p.2x+y=p.

    We must find which options can represent a non-tangent chord and its midpoint (h,k)(h,k)(h,k).

  2. Find intersection of the line with the parabola

    From the line, 2x+y=p  ⟹  x=p−y2.2x+y=p \implies x=\frac{p-y}{2}.2x+y=p⟹x=2p−y​.

    Substitute into the parabola: y2=16⋅p−y2y^2=16\cdot \frac{p-y}{2}y2=16⋅2p−y​ y2=8p−8yy^2=8p-8yy2=8p−8y y2+8y−8p=0.y^2+8y-8p=0.y2+8y−8p=0.

    Let the two intersection points have yyy-coordinates y1,y2y_1,y_2y1​,y2​. Then they are roots of y2+8y−8p=0.y^2+8y-8p=0.y2+8y−8p=0.

  3. Condition that the line is a chord and not a tangent

    For a real chord (two distinct points), the quadratic must have two distinct real roots: Δ=82−4(1)(−8p)=64+32p>0.\Delta = 8^2-4(1)(-8p)=64+32p>0.Δ=82−4(1)(−8p)=64+32p>0. 32(p+2)>0  ⟹  p>−2.32(p+2)>0 \implies p>-2.32(p+2)>0⟹p>−2.

    If p=−2p=-2p=−2, then Δ=0\Delta=0Δ=0 and the line is a tangent, which is not allowed.

  4. Midpoint of the chord in terms of ppp

    Since y1+y2=−8,y_1+y_2=-8,y1​+y2​=−8, the midpoint's yyy-coordinate is k=y1+y22=−4.k=\frac{y_1+y_2}{2}=-4.k=2y1​+y2​​=−4.

    Also, for points on the line 2x+y=p2x+y=p2x+y=p, if midpoint is (h,k)(h,k)(h,k), then it also satisfies the line equation: 2h+k=p.2h+k=p.2h+k=p.

    Using k=−4k=-4k=−4, 2h−4=p2h-4=p2h−4=p h=p+42.h=\frac{p+4}{2}.h=2p+4​.

    So for any such non-tangent chord, k=−4,h=p+42,p>−2.k=-4, \qquad h=\frac{p+4}{2}, \qquad p>-2.k=−4,h=2p+4​,p>−2.

  5. Check each option

    Option A: p=−1, h=1, k=−3p=-1,\ h=1,\ k=-3p=−1, h=1, k=−3

    Required midpoint should have k=−4,k=-4,k=−4, but here k=−3k=-3k=−3.

    Hence A is not possible.

    Option B: p=2, h=3, k=−4p=2,\ h=3,\ k=-4p=2, h=3, k=−4

    Check midpoint formula: h=p+42=2+42=3,h=\frac{p+4}{2}=\frac{2+4}{2}=3,h=2p+4​=22+4​=3, k=−4.k=-4.k=−4. Both match.

    Also, p=2>−2,p=2>-2,p=2>−2, so it is a chord, not a tangent.

    Hence B is possible.

    Option C: p=−2, h=2, k=−4p=-2,\ h=2,\ k=-4p=−2, h=2, k=−4

    Midpoint relation gives h=−2+42=1,h=\frac{-2+4}{2}=1,h=2−2+4​=1, not 222.

    Also, p=−2p=-2p=−2 makes Δ=64+32(−2)=0,\Delta=64+32(-2)=0,Δ=64+32(−2)=0, so the line is a tangent, not a chord.

    Hence C is not possible.

    Option D: p=5, h=4, k=−3p=5,\ h=4,\ k=-3p=5, h=4, k=−3

    Again midpoint must satisfy k=−4,k=-4,k=−4, but here k=−3k=-3k=−3.

    Hence D is not possible.

  6. Final conclusion

    The only possible option is B.\boxed{B}.B​.

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