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Parabola question

2021 · Shift 2 · Q25
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Parabola question

2021 · Shift 2 · Q25

JEE AdvancedMathematicsParabolaMultiple correct+4 / −2
Let E denote the parabola y2 = 8x. Let P = (−-− 2, 4), and let Q and Q' be two distinct points on E such that the lines PQ and PQ' are tangents to E. Let F be the focus of E. Then which of the following statements is(are) TRUE?
  1. A
    The triangle PFQ is a right-angled triangle
  2. B
    The triangle QPQ' is a right-angled triangle
  3. C
    The distance between P and F is 5 2\sqrt 22​
  4. D
    F lies on the line joining Q and Q'
View written solutionFree

Correct answer: A, B, D

Step 1: Analyze the Parabola and Given Points

The equation of the parabola E is given as y2=8xy^2 = 8xy2=8x. This is a standard parabola of the form y2=4axy^2 = 4axy2=4ax. Comparing the two equations, we get 4a = 8, which implies a = 2.

  • The focus F of the parabola is at (a, 0), so F = (2, 0).
  • The vertex is at the origin (0, 0).
  • The equation of the directrix is x = -a, which is x = -2.

The given point is P = (-2, 4).

Step 2: Determine the Location of Point P

By observing the coordinates of point P (-2, 4), we can see that its x-coordinate is -2. This means that point P lies on the directrix of the parabola, x = -2.

This observation is key, as several properties of a parabola are related to its directrix.

Step 3: Evaluate each statement

A: The triangle PFQ is a right-angled triangle

There is a standard property of parabolas: The portion of a tangent between the point of contact and the directrix subtends a right angle at the focus.

In this case, PQ is a tangent to the parabola, where Q is the point of contact and P is a point on the directrix. According to this property, the line segment PQ must subtend a right angle at the focus F. This means that the angle ∠PFQ = 90°. Therefore, the triangle PFQ is a right-angled triangle (right-angled at F).

Alternatively, using slopes: Let the coordinates of Q be (at2,2at)=(2t2,4t)(at^2, 2at) = (2t^2, 4t)(at2,2at)=(2t2,4t). The equation of the tangent at Q is yt=x+at2yt = x + at^2yt=x+at2, which is yt=x+2t2yt = x + 2t^2yt=x+2t2. Since this tangent passes through P(-2, 4), we have 4t=−2+2t24t = -2 + 2t^24t=−2+2t2, which simplifies to t2−2t−1=0t^2 - 2t - 1 = 0t2−2t−1=0. Let t1t_1t1​ be the parameter for point Q. Slope of PF: mPF=(4−0)/(−2−2)=4/(−4)=−1m_{PF} = (4-0)/(-2-2) = 4/(-4) = -1mPF​=(4−0)/(−2−2)=4/(−4)=−1. Slope of FQ: mFQ=(4t1−0)/(2t12−2)=4t1/(2(t12−1))=2t1/(t12−1)m_{FQ} = (4t_1 - 0) / (2t_1^2 - 2) = 4t_1 / (2(t_1^2 - 1)) = 2t_1 / (t_1^2 - 1)mFQ​=(4t1​−0)/(2t12​−2)=4t1​/(2(t12​−1))=2t1​/(t12​−1). For ∠PFQ = 90°, we must have mPF∗mFQ=−1m_{PF} * m_{FQ} = -1mPF​∗mFQ​=−1. (−1)∗(2t1/(t12−1))=−1(-1) * (2t_1 / (t_1^2 - 1)) = -1(−1)∗(2t1​/(t12​−1))=−1 2t1=t12−12t_1 = t_1^2 - 12t1​=t12​−1 t12−2t1−1=0t_1^2 - 2t_1 - 1 = 0t12​−2t1​−1=0. This is the same condition that t1t_1t1​ must satisfy for the tangent to pass through P. Thus, the condition is met. Statement A is TRUE.

B: The triangle QPQ' is a right-angled triangle

Another standard property of parabolas is that tangents drawn from any point on the directrix are perpendicular to each other.

Since P (-2, 4) lies on the directrix, the two tangents PQ and PQ' must be perpendicular. This means the angle ∠QPQ' = 90°. Therefore, the triangle QPQ' is a right-angled triangle (right-angled at P).

Alternatively, using parameters: From the quadratic t2−2t−1=0t^2 - 2t - 1 = 0t2−2t−1=0, let the roots be t1t_1t1​ and t2t_2t2​, corresponding to points Q and Q'. The slope of the tangent at (2t2,4t)(2t^2, 4t)(2t2,4t) is 1/t. So the slopes of tangents PQ and PQ' are m1=1/t1m_1 = 1/t_1m1​=1/t1​ and m2=1/t2m_2 = 1/t_2m2​=1/t2​. The product of the slopes is m1m2=(1/t1)(1/t2)=1/(t1t2)m_1 m_2 = (1/t_1)(1/t_2) = 1/(t_1 t_2)m1​m2​=(1/t1​)(1/t2​)=1/(t1​t2​). From Vieta's formulas for the quadratic, the product of roots t1t2=−1t_1 t_2 = -1t1​t2​=−1. So, m1m2=−1m_1 m_2 = -1m1​m2​=−1. The tangents are perpendicular. Statement B is TRUE.

C: The distance between P and F is 5√2

The coordinates are P (-2, 4) and F (2, 0). We use the distance formula: PF=(x2−x1)2+(y2−y1)2PF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}PF=(x2​−x1​)2+(y2​−y1​)2​ PF=(2−(−2))2+(0−4)2PF = \sqrt{(2 - (-2))^2 + (0 - 4)^2}PF=(2−(−2))2+(0−4)2​ PF=(4)2+(−4)2PF = \sqrt{(4)^2 + (-4)^2}PF=(4)2+(−4)2​ PF=16+16=32=16×2=42PF = \sqrt{16 + 16} = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}PF=16+16​=32​=16×2​=42​.

The statement says the distance is 525\sqrt{2}52​, which is incorrect. Statement C is FALSE.

D: F lies on the line joining Q and Q'

A third standard property of parabolas is that the chord of contact of tangents drawn from any point on the directrix is a focal chord (i.e., it passes through the focus).

The line joining Q and Q' is the chord of contact for tangents drawn from point P. Since P lies on the directrix, the chord of contact QQ' must pass through the focus F. Therefore, F lies on the line joining Q and Q'.

Alternatively, using chord of contact equation: The equation of the chord of contact from an external point (x0,y0)(x_0, y_0)(x0​,y0​) to the parabola y2=4axy^2 = 4axy2=4ax is yy0=2a(x+x0)yy_0 = 2a(x + x_0)yy0​=2a(x+x0​). For P (-2, 4) and a=2, the equation of QQ' is: y(4) = 2(2)(x + (-2)) 4y = 4(x - 2) y = x - 2. To check if the focus F (2, 0) lies on this line, substitute its coordinates into the equation: 0 = 2 - 2, which is 0 = 0. The condition is satisfied. Statement D is TRUE.

Conclusion

Based on the analysis, statements A, B, and D are true, while statement C is false.

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