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Ellipse question

2009 · Shift 1 · Q30
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  5. /2009 · Shift 1 · Q30

Ellipse question

2009 · Shift 1 · Q30

JEE AdvancedMathematicsEllipseMCQ+3 / −1

Match the conics in Column I with the statements/expressions in Column II :

Column I Column II
(A) Circle (P) The locus of the point (h,kh,kh,k) for which the line hx+ky=1hx+ky=1hx+ky=1 touches the circle x2+y2=4x^2+y^2=4x2+y2=4.
(B) Parabola (Q) Points z in the complex plane satisfying ∣z+2∣−∣z−2∣=±3|z+2|-|z-2|=\pm3∣z+2∣−∣z−2∣=±3.
(C) Ellipse (R) Points of the conic have parametric representation x=3(1−t21+t2),y=2t1+t2x = \sqrt 3 \left( {{{1 - {t^2}} \over {1 + {t^2}}}} \right),y = {{2t} \over {1 + {t^2}}}x=3​(1+t21−t2​),y=1+t22t​
(D) Hyperbola (S) The eccentricity of the conic lies in the interval 1≤x≤∞1 \le x \le \infty1≤x≤∞.
(T) Points z in the complex plane satisfying Reolimits(z+1)2=∣z∣2+1{\mathop{\rm Re} olimits} {(z + 1)^2} = |z{|^2} + 1Reolimits(z+1)2=∣z∣2+1.

  1. A
    (A) →\to→(P); (B) →\to→(S), (T); (C) →\to→(R); (D) →\to→(R), (S)
  2. B
    (A) →\to→(P); (B) →\to→(S), (T); (C) →\to→(R); (D) →\to→(Q), (S)
  3. C
    (A) →\to→(P); (B) →\to→(S), (T); (C) →\to→(S); (D) →\to→(R), (S)
  4. D
    (A) →\to→(P); (B) →\to→(P), (T); (C) →\to→(R); (D) →\to→(Q), (S)
View written solutionFree

Correct answer: B

To solve this matching problem, we need to analyze each statement/expression in Column II and determine the type of conic section it represents. Then we can match it with the conic names in Column I.

Analyzing Column II

Statement (P): The locus of the point (h,kh,kh,k) for which the line hx+ky=1hx+ky=1hx+ky=1 touches the circle x2+y2=4x^2+y^2=4x2+y2=4.

  1. The equation of the given circle is x2+y2=4x^2+y^2=4x2+y2=4. This is a circle with its center at the origin (0,0)(0,0)(0,0) and radius r=2r=2r=2.
  2. The equation of the line is hx+ky=1hx+ky=1hx+ky=1, which can be rewritten as hx+ky−1=0hx+ky-1=0hx+ky−1=0.
  3. The condition for a line to be tangent to a circle is that the perpendicular distance from the center of the circle to the line is equal to the radius of the circle.
  4. The perpendicular distance ddd from a point (x1,y1)(x_1, y_1)(x1​,y1​) to a line Ax+By+C=0Ax+By+C=0Ax+By+C=0 is given by d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}d=A2+B2​∣Ax1​+By1​+C∣​.
  5. Here, (x1,y1)=(0,0)(x_1, y_1) = (0,0)(x1​,y1​)=(0,0), A=hA=hA=h, B=kB=kB=k, and C=−1C=-1C=−1. The radius is r=2r=2r=2.
  6. So, we must have: d=∣h(0)+k(0)−1∣h2+k2=2d = \frac{|h(0)+k(0)-1|}{\sqrt{h^2+k^2}} = 2d=h2+k2​∣h(0)+k(0)−1∣​=2 1h2+k2=2\frac{1}{\sqrt{h^2+k^2}} = 2h2+k2​1​=2
  7. Squaring both sides gives: 1h2+k2=4  ⟹  h2+k2=14\frac{1}{h^2+k^2} = 4 \implies h^2+k^2 = \frac{1}{4}h2+k21​=4⟹h2+k2=41​
  8. The locus of the point (h,k)(h,k)(h,k) is x2+y2=(1/2)2x^2+y^2 = (1/2)^2x2+y2=(1/2)2. This is the equation of a circle.
  9. Therefore, (A) →\to→ (P).

Statement (Q): Points z in the complex plane satisfying ∣z+2∣−∣z−2∣=±3|z+2|-|z-2|=\pm3∣z+2∣−∣z−2∣=±3.

  1. Let z=x+iyz=x+iyz=x+iy. The equation represents the locus of a point zzz in the complex plane.
  2. ∣z+2∣|z+2|∣z+2∣ is the distance of the point zzz from the point (−2,0)(-2,0)(−2,0). Let's call this point F1F_1F1​. So PF1=∣z−(−2)∣=∣z+2∣PF_1 = |z-(-2)| = |z+2|PF1​=∣z−(−2)∣=∣z+2∣.
  3. ∣z−2∣|z-2|∣z−2∣ is the distance of the point zzz from the point (2,0)(2,0)(2,0). Let's call this point F2F_2F2​. So PF2=∣z−2∣PF_2 = |z-2|PF2​=∣z−2∣.
  4. The equation is PF1−PF2=±3PF_1 - PF_2 = \pm3PF1​−PF2​=±3. This means the difference of the distances of any point on the locus from two fixed points (foci) F1(−2,0)F_1(-2,0)F1​(−2,0) and F2(2,0)F_2(2,0)F2​(2,0) is a constant, which is 333.
  5. This is the definition of a hyperbola.
  6. The distance between foci is 2c=∣2−(−2)∣=42c = |2 - (-2)| = 42c=∣2−(−2)∣=4, so c=2c=2c=2. The constant difference is 2a=32a=32a=3, so a=3/2a=3/2a=3/2.
  7. Since c>ac>ac>a (2>3/22 > 3/22>3/2), it is indeed a hyperbola.
  8. Therefore, (D) →\to→ (Q).

Statement (R): Points of the conic have parametric representation x=3(1−t21+t2),y=2t1+t2x = \sqrt 3 \left( {{{1 - {t^2}} \over {1 + {t^2}}}} \right),y = {{2t} \over {1 + {t^2}}}x=3​(1+t21−t2​),y=1+t22t​.

  1. We can use the standard substitution t=tan⁡(θ/2)t = \tan(\theta/2)t=tan(θ/2). This gives cos⁡θ=1−t21+t2\cos\theta = \frac{1-t^2}{1+t^2}cosθ=1+t21−t2​ and sin⁡θ=2t1+t2\sin\theta = \frac{2t}{1+t^2}sinθ=1+t22t​.
  2. Substituting these into the given parametric equations: x=3cos⁡θ  ⟹  x3=cos⁡θx = \sqrt{3} \cos\theta \implies \frac{x}{\sqrt{3}} = \cos\thetax=3​cosθ⟹3​x​=cosθ y=sin⁡θy = \sin\thetay=sinθ
  3. To find the Cartesian equation, we can eliminate the parameter θ\thetaθ using the identity cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1cos2θ+sin2θ=1. (x3)2+y2=1\left(\frac{x}{\sqrt{3}}\right)^2 + y^2 = 1(3​x​)2+y2=1 x23+y21=1\frac{x^2}{3} + \frac{y^2}{1} = 13x2​+1y2​=1
  4. This is the standard equation of an ellipse centered at the origin, with semi-major axis a=3a=\sqrt{3}a=3​ and semi-minor axis b=1b=1b=1.
  5. Therefore, (C) →\to→ (R).

Statement (S): The eccentricity of the conic lies in the interval 1≤e<∞1 \le e < \infty1≤e<∞.

  1. Let's recall the eccentricities (eee) for different conic sections:
    • Circle: e=0e=0e=0
    • Ellipse: 0<e<10 < e < 10<e<1
    • Parabola: e=1e=1e=1
    • Hyperbola: e>1e > 1e>1
  2. The interval 1≤e<∞1 \le e < \infty1≤e<∞ includes e=1e=1e=1 and all values greater than 1. This corresponds to both Parabola and Hyperbola.
  3. Therefore, (B) →\to→ (S) and (D) →\to→ (S).

Statement (T): Points z in the complex plane satisfying Re(z+1)2=∣z∣2+1{\mathop{\rm Re}} {(z + 1)^2} = |z{|^2} + 1Re(z+1)2=∣z∣2+1.

  1. Let z=x+iyz = x+iyz=x+iy.
  2. The left-hand side (LHS) is Re(z+1)2=Re((x+1)+iy)2{\mathop{\rm Re}} {(z + 1)^2} = {\mathop{\rm Re}} {((x+1)+iy)^2}Re(z+1)2=Re((x+1)+iy)2. LHS =Re((x+1)2−y2+2i(x+1)y)=(x+1)2−y2=x2+2x+1−y2= {\mathop{\rm Re}} {( (x+1)^2 - y^2 + 2i(x+1)y )} = (x+1)^2 - y^2 = x^2+2x+1-y^2=Re((x+1)2−y2+2i(x+1)y)=(x+1)2−y2=x2+2x+1−y2.
  3. The right-hand side (RHS) is ∣z∣2+1=∣x+iy∣2+1=x2+y2+1|z|^2 + 1 = |x+iy|^2 + 1 = x^2+y^2+1∣z∣2+1=∣x+iy∣2+1=x2+y2+1.
  4. Equating LHS and RHS: x2+2x+1−y2=x2+y2+1x^2+2x+1-y^2 = x^2+y^2+1x2+2x+1−y2=x2+y2+1 2x−y2=y22x - y^2 = y^22x−y2=y2 2x=2y22x = 2y^22x=2y2 y2=xy^2 = xy2=x
  5. This is the standard equation of a parabola with its vertex at the origin and opening to the right.
  6. Therefore, (B) →\to→ (T).

Matching Summary

  • (A) Circle matches with (P).
  • (B) Parabola matches with (S) and (T).
  • (C) Ellipse matches with (R).
  • (D) Hyperbola matches with (Q) and (S).

This gives the following mapping:

  • (A) →\to→ (P)
  • (B) →\to→ (S), (T)
  • (C) →\to→ (R)
  • (D) →\to→ (Q), (S)

Comparing with Options

  • A: (A) →\to→(P); (B) →\to→(S), (T); (C) →\to→(R); (D) →\to→(R), (S) - Incorrect because (D) maps to (Q), not (R).
  • B: (A) →\to→(P); (B) →\to→(S), (T); (C) →\to→(R); (D) →\to→(Q), (S) - This option correctly matches all our findings.
  • C: (A) →\to→(P); (B) →\to→(S), (T); (C) →\to→(S); (D) →\to→(R), (S) - Incorrect because (C) does not map to (S) and (D) does not map to (R).
  • D: (A) →\to→(P); (B) →\to→(P), (T); (C) →\to→(R); (D) →\to→(Q), (S) - Incorrect because (B) does not map to (P).

The correct option is B.

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