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Parabola question

2013 · Shift 2 · Q32
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Parabola question

2013 · Shift 2 · Q32

JEE AdvancedMathematicsParabolaMCQ+4 / −1
Let PQPQPQ be a focal chord of the parabola y2=4ax{y^2} = 4axy2=4ax. The tangents to the parabola at PPP and QQQ meet at a point lying on the line y=2x+ay=2x+ay=2x+a, a>0a\gt 0a>0. Length of chord PQPQPQ is
  1. A
    7a7a7a
  2. B
    5a5a5a
  3. C
    2a2a2a
  4. D
    3a3a3a
View written solutionFree

Correct answer: B

Step-by-step Derivation

  1. Parametric Coordinates on the Parabola The equation of the parabola is y2=4axy^2 = 4axy2=4ax. Let the endpoints of the focal chord PQPQPQ be P(at12,2at1)P(at_1^2, 2at_1)P(at12​,2at1​) and Q(at22,2at2)Q(at_2^2, 2at_2)Q(at22​,2at2​), where t1t_1t1​ and t2t_2t2​ are parameters.

  2. Condition for a Focal Chord A chord is a focal chord if it passes through the focus, S(a,0)S(a,0)S(a,0). The condition for the chord connecting points with parameters t1t_1t1​ and t2t_2t2​ to be a focal chord is: t1t2=−1t_1 t_2 = -1t1​t2​=−1

  3. Point of Intersection of Tangents The equation of the tangent to the parabola y2=4axy^2 = 4axy2=4ax at a point (at2,2at)(at^2, 2at)(at2,2at) is given by ty=x+at2ty = x + at^2ty=x+at2. The tangent at PPP is: t1y=x+at12t_1y = x + at_1^2t1​y=x+at12​. The tangent at QQQ is: t2y=x+at22t_2y = x + at_2^2t2​y=x+at22​. The point of intersection of these two tangents is a standard result, given by the coordinates (at1t2,a(t1+t2))(a t_1 t_2, a(t_1 + t_2))(at1​t2​,a(t1​+t2​)).

  4. Applying the Given Condition Since PQPQPQ is a focal chord, we use the condition t1t2=−1t_1t_2 = -1t1​t2​=−1. The point of intersection becomes (−a,a(t1+t2))(-a, a(t_1 + t_2))(−a,a(t1​+t2​)). We are given that this point of intersection lies on the line y=2x+ay = 2x + ay=2x+a. Substituting the coordinates of the intersection point into the line's equation: y=2x+ay = 2x + ay=2x+a a(t1+t2)=2(−a)+aa(t_1 + t_2) = 2(-a) + aa(t1​+t2​)=2(−a)+a a(t1+t2)=−2a+aa(t_1 + t_2) = -2a + aa(t1​+t2​)=−2a+a a(t1+t2)=−aa(t_1 + t_2) = -aa(t1​+t2​)=−a Since a>0a > 0a>0, we can divide by aaa: t1+t2=−1t_1 + t_2 = -1t1​+t2​=−1

  5. Calculating the Length of the Focal Chord The length of a focal chord PQPQPQ can be found by summing the focal distances of its endpoints, SP+SQSP + SQSP+SQ. The focal distance of a point (x,y)(x,y)(x,y) on the parabola y2=4axy^2 = 4axy2=4ax is x+ax+ax+a. For point P(at12,2at1)P(at_1^2, 2at_1)P(at12​,2at1​), the focal distance SP=at12+aSP = at_1^2 + aSP=at12​+a. For point Q(at22,2at2)Q(at_2^2, 2at_2)Q(at22​,2at2​), the focal distance SQ=at22+aSQ = at_2^2 + aSQ=at22​+a. The length of the chord PQPQPQ is: L=SP+SQ=(at12+a)+(at22+a)L = SP + SQ = (at_1^2 + a) + (at_2^2 + a)L=SP+SQ=(at12​+a)+(at22​+a) L=a(t12+t22+2)L = a(t_1^2 + t_2^2 + 2)L=a(t12​+t22​+2) We can express t12+t22t_1^2 + t_2^2t12​+t22​ in terms of t1+t2t_1+t_2t1​+t2​ and t1t2t_1t_2t1​t2​: t12+t22=(t1+t2)2−2t1t2t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2t12​+t22​=(t1​+t2​)2−2t1​t2​ From our previous steps, we know t1+t2=−1t_1 + t_2 = -1t1​+t2​=−1 and t1t2=−1t_1t_2 = -1t1​t2​=−1. Substituting these values: t12+t22=(−1)2−2(−1)=1+2=3t_1^2 + t_2^2 = (-1)^2 - 2(-1) = 1 + 2 = 3t12​+t22​=(−1)2−2(−1)=1+2=3 Now, substitute this result back into the formula for the length LLL: L=a(3+2)=5aL = a(3 + 2) = 5aL=a(3+2)=5a

    Alternative Method using Distance Formula: The length of the chord PQPQPQ can also be calculated using the distance formula: PQ2=(at12−at22)2+(2at1−2at2)2PQ^2 = (at_1^2 - at_2^2)^2 + (2at_1 - 2at_2)^2PQ2=(at12​−at22​)2+(2at1​−2at2​)2 PQ2=a2(t12−t22)2+4a2(t1−t2)2PQ^2 = a^2(t_1^2 - t_2^2)^2 + 4a^2(t_1 - t_2)^2PQ2=a2(t12​−t22​)2+4a2(t1​−t2​)2 PQ2=a2[(t1−t2)(t1+t2)]2+4a2(t1−t2)2PQ^2 = a^2[(t_1-t_2)(t_1+t_2)]^2 + 4a^2(t_1 - t_2)^2PQ2=a2[(t1​−t2​)(t1​+t2​)]2+4a2(t1​−t2​)2 PQ2=a2(t1−t2)2[(t1+t2)2+4]PQ^2 = a^2(t_1-t_2)^2[(t_1+t_2)^2 + 4]PQ2=a2(t1​−t2​)2[(t1​+t2​)2+4] We need to find (t1−t2)2(t_1-t_2)^2(t1​−t2​)2: (t1−t2)2=(t1+t2)2−4t1t2=(−1)2−4(−1)=1+4=5(t_1-t_2)^2 = (t_1+t_2)^2 - 4t_1t_2 = (-1)^2 - 4(-1) = 1+4 = 5(t1​−t2​)2=(t1​+t2​)2−4t1​t2​=(−1)2−4(−1)=1+4=5 Substituting the values: PQ2=a2(5)[(−1)2+4]=a2(5)(1+4)=25a2PQ^2 = a^2(5)[(-1)^2 + 4] = a^2(5)(1+4) = 25a^2PQ2=a2(5)[(−1)2+4]=a2(5)(1+4)=25a2 PQ = oxed{5a}

  6. Conclusion The length of the focal chord PQPQPQ is 5a5a5a. This corresponds to option B.

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