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Question: Two curves C and D are defined as $C = 17y^2x^9 - 71x^5 + 25x^4 - 3x^2 + 4y^2 = 0$ $D = 18y^2x^9 - 7...

Two curves C and D are defined as C=17y2x9−71x5+25x4−3x2+4y2=0C = 17y^2x^9 - 71x^5 + 25x^4 - 3x^2 + 4y^2 = 0 D=18y2x9−72x5+26x4+3x3−4xy2=0D = 18y^2x^9 - 72x^5 + 26x^4 + 3x^3 - 4xy^2 = 0 Let T1,T2,T3,...,Tm(m∈N)T_1, T_2, T_3, ..., T_m (m \in N) be the tangents to curve C at origin and N1,N2,...,Nn(n∈N)N_1, N_2, ..., N_n (n \in N) be the normals to curve D at origin. Then, which of the following options can become equal to the angle between lines LiL_i and Nj(i,j∈N,i≤m,j≤n)N_j (i, j \in N, i \leq m, j \leq n)?

A

tan−1(32)tan^{-1}(\frac{\sqrt{3}}{2})

B

π4−tan−1(7−43)\frac{\pi}{4} - tan^{-1}(7 - 4\sqrt{3})

C

π2\frac{\pi}{2}

D

tan−1(143)tan^{-1}(\frac{1}{4\sqrt{3}})

Answer

A, B, C, D

Explanation

Solution

The tangents to curve C at the origin are found by equating the lowest degree terms to zero: −3x2+4y2=0-3x^2 + 4y^2 = 0, which gives y=±32xy = \pm \frac{\sqrt{3}}{2}x. The slopes of these tangents are mTC=±32m_{T_C} = \pm \frac{\sqrt{3}}{2}.

The tangents to curve D at the origin are found by equating the lowest degree terms to zero: 3x3−4xy2=03x^3 - 4xy^2 = 0, which gives x(3x2−4y2)=0x(3x^2 - 4y^2) = 0. The tangents are x=0x=0 and y=±32xy = \pm \frac{\sqrt{3}}{2}x.

The normals to curve D at the origin are:

  1. For tangent x=0x=0 (y-axis), the normal is y=0y=0 (x-axis) with slope mN1=0m_{N_1} = 0.
  2. For tangent y=32xy = \frac{\sqrt{3}}{2}x, the normal has slope mN2=−13/2=−23m_{N_2} = -\frac{1}{\sqrt{3}/2} = -\frac{2}{\sqrt{3}}.
  3. For tangent y=−32xy = -\frac{\sqrt{3}}{2}x, the normal has slope mN3=−1−3/2=23m_{N_3} = -\frac{1}{-\sqrt{3}/2} = \frac{2}{\sqrt{3}}. The slopes of the normals to curve D are mND∈{0,−23,23}m_{N_D} \in \{0, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\}.

The angle θ\theta between a tangent to C (mTCm_{T_C}) and a normal to D (mNDm_{N_D}) is given by tan⁡θ=∣mTC−mND1+mTCmND∣\tan \theta = |\frac{m_{T_C} - m_{N_D}}{1 + m_{T_C} m_{N_D}}|.

Possible angles:

  1. mTC=32m_{T_C} = \frac{\sqrt{3}}{2}, mND=0m_{N_D} = 0: tan⁡θ=∣3/2−01+0∣=32\tan \theta = |\frac{\sqrt{3}/2 - 0}{1 + 0}| = \frac{\sqrt{3}}{2}. So, θ=tan⁡−1(32)\theta = \tan^{-1}(\frac{\sqrt{3}}{2}). (Option A)
  2. mTC=32m_{T_C} = \frac{\sqrt{3}}{2}, mND=−23m_{N_D} = -\frac{2}{\sqrt{3}}: 1+mTCmND=1+(32)(−23)=1−1=01 + m_{T_C} m_{N_D} = 1 + (\frac{\sqrt{3}}{2})(-\frac{2}{\sqrt{3}}) = 1 - 1 = 0. The lines are perpendicular, so θ=π2\theta = \frac{\pi}{2}. (Option C)
  3. mTC=32m_{T_C} = \frac{\sqrt{3}}{2}, mND=23m_{N_D} = \frac{2}{\sqrt{3}}: tan⁡θ=∣3/2−2/31+(3/2)(2/3)∣=∣(3−4)/(23)1+1∣=143\tan \theta = |\frac{\sqrt{3}/2 - 2/\sqrt{3}}{1 + (\sqrt{3}/2)(2/\sqrt{3})}| = |\frac{(3-4)/(2\sqrt{3})}{1+1}| = \frac{1}{4\sqrt{3}}. So, θ=tan⁡−1(143)\theta = \tan^{-1}(\frac{1}{4\sqrt{3}}). (Option D)

For Option B, let θB=π4−tan⁡−1(7−43)\theta_B = \frac{\pi}{4} - \tan^{-1}(7 - 4\sqrt{3}). Let α=tan⁡−1(7−43)\alpha = \tan^{-1}(7 - 4\sqrt{3}). tan⁡(θB)=tan⁡(π4−α)=1−tan⁡α1+tan⁡α=1−(7−43)1+(7−43)=−6+438−43=2(23−3)4(2−3)=23−32(2−3)\tan(\theta_B) = \tan(\frac{\pi}{4} - \alpha) = \frac{1 - \tan \alpha}{1 + \tan \alpha} = \frac{1 - (7 - 4\sqrt{3})}{1 + (7 - 4\sqrt{3})} = \frac{-6 + 4\sqrt{3}}{8 - 4\sqrt{3}} = \frac{2(2\sqrt{3} - 3)}{4(2 - \sqrt{3})} = \frac{2\sqrt{3} - 3}{2(2 - \sqrt{3})}. Rationalizing the denominator: 23−32(2−3)×2+32+3=(23−3)(2+3)2(4−3)=43+6−6−332=32\frac{2\sqrt{3} - 3}{2(2 - \sqrt{3})} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{(2\sqrt{3} - 3)(2 + \sqrt{3})}{2(4 - 3)} = \frac{4\sqrt{3} + 6 - 6 - 3\sqrt{3}}{2} = \frac{\sqrt{3}}{2}. Thus, θB=tan⁡−1(32)\theta_B = \tan^{-1}(\frac{\sqrt{3}}{2}), which is equal to Option A.

Since all calculated angles match options A, C, and D, and option B is equivalent to option A, all options can become equal to the angle between the lines.