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Ellipse question

2009 · Shift 1 · Q31
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  1. PYQ Library
  2. /JEE Advanced
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  4. /Ellipse
  5. /2009 · Shift 1 · Q31

Ellipse question

2009 · Shift 1 · Q31

JEE AdvancedMathematicsEllipseMultiple correct+4 / −2
In a triangle ABCABCABC with fixed base BCBCBC, the vertex AAA moves such that cos⁡ B+cos⁡ C=4sin⁡2A2.\cos \,B + \cos \,C = 4{\sin ^2}{A \over 2}.cosB+cosC=4sin22A​. If a,ba, ba,b and ccc denote the lengths of the sides of the triangle opposite to the angles A,BA, BA,B and CCC, respectively, then
  1. A
    b+c=4ab+c=4ab+c=4a
  2. B
    b+c=2ab+c=2ab+c=2a
  3. C
    locus of point AAA is an ellipse
  4. D
    locus of point AAA is a pair of straight lines
View written solutionFree

Correct answer: B, C

  1. Use the cosine rule expressions for cos⁡B\cos BcosB and cos⁡C\cos CcosC

In triangle ABCABCABC with sides opposite A,B,CA,B,CA,B,C equal to a,b,ca,b,ca,b,c respectively,

cos⁡B=a2+c2−b22ac,cos⁡C=a2+b2−c22ab.\cos B=\frac{a^2+c^2-b^2}{2ac}, \qquad \cos C=\frac{a^2+b^2-c^2}{2ab}.cosB=2aca2+c2−b2​,cosC=2aba2+b2−c2​.

So,

cos⁡B+cos⁡C=a2+c2−b22ac+a2+b2−c22ab.\cos B+\cos C =\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}.cosB+cosC=2aca2+c2−b2​+2aba2+b2−c2​.

Taking LCM 2abc2abc2abc,

cos⁡B+cos⁡C=b(a2+c2−b2)+c(a2+b2−c2)2abc.\cos B+\cos C =\frac{b(a^2+c^2-b^2)+c(a^2+b^2-c^2)}{2abc}.cosB+cosC=2abcb(a2+c2−b2)+c(a2+b2−c2)​.

Simplify the numerator:

=a2(b+c)+bc(b+c)−(b3+c3)2abc.=\frac{a^2(b+c)+bc(b+c)-\big(b^3+c^3\big)}{2abc}.=2abca2(b+c)+bc(b+c)−(b3+c3)​.

Now,

b3+c3=(b+c)(b2−bc+c2),b^3+c^3=(b+c)(b^2-bc+c^2),b3+c3=(b+c)(b2−bc+c2),

hence

bc(b+c)−(b3+c3)=(b+c)[bc−(b2−bc+c2)]=−(b+c)(b−c)2.bc(b+c)-(b^3+c^3) =(b+c)\big[bc-(b^2-bc+c^2)\big] =-(b+c)(b-c)^2.bc(b+c)−(b3+c3)=(b+c)[bc−(b2−bc+c2)]=−(b+c)(b−c)2.

Thus,

cos⁡B+cos⁡C=(b+c)(a2−(b−c)2)2abc.\cos B+\cos C =\frac{(b+c)\big(a^2-(b-c)^2\big)}{2abc}.cosB+cosC=2abc(b+c)(a2−(b−c)2)​.

But

a2−(b−c)2=(a−b+c)(a+b−c).a^2-(b-c)^2=(a-b+c)(a+b-c).a2−(b−c)2=(a−b+c)(a+b−c).

Using Heron-type identity,

a2−(b−c)2=4bccos⁡2A2.a^2-(b-c)^2=4bc\cos^2\frac A2.a2−(b−c)2=4bccos22A​.

Therefore,

cos⁡B+cos⁡C=(b+c)⋅4bccos⁡2A22abc=2(b+c)acos⁡2A2.\cos B+\cos C =\frac{(b+c)\cdot 4bc\cos^2\frac A2}{2abc} =\frac{2(b+c)}{a}\cos^2\frac A2.cosB+cosC=2abc(b+c)⋅4bccos22A​​=a2(b+c)​cos22A​.
  1. Use the given condition

Given,

cos⁡B+cos⁡C=4sin⁡2A2.\cos B+\cos C=4\sin^2\frac A2.cosB+cosC=4sin22A​.

Substitute the above result:

2(b+c)acos⁡2A2=4sin⁡2A2.\frac{2(b+c)}{a}\cos^2\frac A2=4\sin^2\frac A2.a2(b+c)​cos22A​=4sin22A​.

So,

b+cacos⁡2A2=2sin⁡2A2.\frac{b+c}{a}\cos^2\frac A2=2\sin^2\frac A2.ab+c​cos22A​=2sin22A​.

At this point, it is better to use a standard identity directly:

cos⁡B+cos⁡C=2cos⁡B+C2cos⁡B−C2.\cos B+\cos C=2\cos\frac{B+C}{2}\cos\frac{B-C}{2}.cosB+cosC=2cos2B+C​cos2B−C​.

Since B+C=π−AB+C=\pi-AB+C=π−A,

cos⁡B+C2=cos⁡π−A2=sin⁡A2.\cos\frac{B+C}{2}=\cos\frac{\pi-A}{2}=\sin\frac A2.cos2B+C​=cos2π−A​=sin2A​.

Hence,

cos⁡B+cos⁡C=2sin⁡A2cos⁡B−C2.\cos B+\cos C=2\sin\frac A2\cos\frac{B-C}{2}.cosB+cosC=2sin2A​cos2B−C​.

Given this equals 4sin⁡2A24\sin^2\frac A24sin22A​, so

2sin⁡A2cos⁡B−C2=4sin⁡2A2.2\sin\frac A2\cos\frac{B-C}{2}=4\sin^2\frac A2.2sin2A​cos2B−C​=4sin22A​.

For non-degenerate triangle, sin⁡A2≠0\sin\frac A2\neq 0sin2A​=0, therefore

cos⁡B−C2=2sin⁡A2.\cos\frac{B-C}{2}=2\sin\frac A2.cos2B−C​=2sin2A​.

Now use the identity

b+c=2acos⁡B−C2sec⁡A2.b+c=2a\cos\frac{B-C}{2}\sec\frac A2.b+c=2acos2B−C​sec2A​.

Substituting cos⁡B−C2=2sin⁡A2\cos\frac{B-C}{2}=2\sin\frac A2cos2B−C​=2sin2A​,

b+c=2a⋅2sin⁡A2sec⁡A2=4atan⁡A2.b+c=2a\cdot 2\sin\frac A2\sec\frac A2 =4a\tan\frac A2.b+c=2a⋅2sin2A​sec2A​=4atan2A​.

This route is messy. So let us use the standard half-angle side formulas:

sin⁡B2=(s−a)(s−c)ac,sin⁡C2=(s−a)(s−b)ab.\sin\frac B2=\sqrt{\frac{(s-a)(s-c)}{ac}}, \qquad \sin\frac C2=\sqrt{\frac{(s-a)(s-b)}{ab}}.sin2B​=ac(s−a)(s−c)​​,sin2C​=ab(s−a)(s−b)​​.

A much simpler route is to use

cos⁡B+cos⁡C=2cos⁡B+C2cos⁡B−C2=2sin⁡A2cos⁡B−C2.\cos B+\cos C=2\cos\frac{B+C}{2}\cos\frac{B-C}{2} =2\sin\frac A2\cos\frac{B-C}{2}.cosB+cosC=2cos2B+C​cos2B−C​=2sin2A​cos2B−C​.

Also,

b+c2a=sin⁡B+sin⁡C2sin⁡A=2sin⁡B+C2cos⁡B−C22⋅2sin⁡A2cos⁡A2=2cos⁡A2cos⁡B−C24sin⁡A2cos⁡A2=cos⁡B−C22sin⁡A2.\frac{b+c}{2a}=\frac{\sin B+\sin C}{2\sin A} =\frac{2\sin\frac{B+C}{2}\cos\frac{B-C}{2}}{2\cdot 2\sin\frac A2\cos\frac A2} =\frac{2\cos\frac A2\cos\frac{B-C}{2}}{4\sin\frac A2\cos\frac A2} =\frac{\cos\frac{B-C}{2}}{2\sin\frac A2}.2ab+c​=2sinAsinB+sinC​=2⋅2sin2A​cos2A​2sin2B+C​cos2B−C​​=4sin2A​cos2A​2cos2A​cos2B−C​​=2sin2A​cos2B−C​​.

Thus,

cos⁡B−C2=b+casin⁡A2.\cos\frac{B-C}{2}=\frac{b+c}{a}\sin\frac A2.cos2B−C​=ab+c​sin2A​.

Substitute into

cos⁡B+cos⁡C=2sin⁡A2cos⁡B−C2:\cos B+\cos C=2\sin\frac A2\cos\frac{B-C}{2}:cosB+cosC=2sin2A​cos2B−C​: cos⁡B+cos⁡C=2sin⁡A2(b+casin⁡A2)=2(b+c)asin⁡2A2.\cos B+\cos C =2\sin\frac A2\left(\frac{b+c}{a}\sin\frac A2\right) =\frac{2(b+c)}{a}\sin^2\frac A2.cosB+cosC=2sin2A​(ab+c​sin2A​)=a2(b+c)​sin22A​.

Given this equals 4sin⁡2A24\sin^2\frac A24sin22A​, hence

2(b+c)a=4⇒b+c=2a.\frac{2(b+c)}{a}=4 \quad\Rightarrow\quad b+c=2a.a2(b+c)​=4⇒b+c=2a.

So Option B is correct and Option A is false.

  1. Interpret the locus of AAA

Here a=BCa=BCa=BC is fixed because base BCBCBC is fixed. Also,

b=CA,c=AB.b=CA,\qquad c=AB.b=CA,c=AB.

The condition found is

AB+AC=2 BC.AB+AC=2\,BC.AB+AC=2BC.

This means the sum of distances of moving point AAA from two fixed points BBB and CCC is constant. That is precisely the definition of an ellipse with foci at BBB and CCC.

Hence Option C is correct.

A pair of straight lines would arise from a condition like difference of distances being constant in a limiting/degenerate form, not here. So Option D is false.

  1. Final evaluation of options
  • A: b+c=4ab+c=4ab+c=4a — False
  • B: b+c=2ab+c=2ab+c=2a — True
  • C: locus of point AAA is an ellipse — True
  • D: locus of point AAA is a pair of straight lines — False

Therefore, the correct options are

B, C\boxed{\text{B, C}}B, C​
  1. Comparison with stored correct answer

Stored correct answer: B, C\text{B, C}B, C

This matches the derived answer.

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