Jee Mains

For all $\mathrm{z} \in \mathrm{C}$ on the curve $\mathrm{C}_{1}:|\mathrm{z}|=4$, let the locus of the point $\mathrm{z}+\frac{1}{\mathrm{z}}$ be the curve $\mathrm{C}_{2}$. Then

Tagged: ,

Log In Register Lost Password
Viewing 3 posts - 1 through 3 (of 3 total)
  • Author
    Posts
  • 30 October, 2026 at 8:11 am #12547
    UPLIFTJEE
    Keymaster

    For all $\mathrm{z} \in \mathrm{C}$ on the curve $\mathrm{C}_{1}:|\mathrm{z}|=4$, let the locus of the point $\mathrm{z}+\frac{1}{\mathrm{z}}$ be the curve $\mathrm{C}_{2}$. Then [JEE (Main) 31 Jan 2023 S I]
    (A) the curves $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ intersect at 2 points
    (B) the curves $\mathrm{C}_{1}$ lies inside $\mathrm{C}_{2}$
    (C) the curves $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ intersect at 4 points
    (D) the curves $\mathrm{C}_{2}$ lies inside $\mathrm{C}_{1}$

    30 October, 2026 at 8:11 am #12548
    UPLIFTJEE
    Keymaster

    Solution

    Let $\mathrm{w}=\mathrm{z}+\frac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\frac{1}{4} 4 \mathrm{e}^{-\mathrm{i} \theta} \Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta$
    So locus of w is ellipse $\frac{x^{2}}{\left(\frac{17}{4}\right)^{2}}+\frac{y^{2}}{\left(\frac{15}{4}\right)^{2}}=1$ Locus of z is circle $x^{2}+y^{2}=16$ So intersect at 4 points

    30 October, 2026 at 8:11 am #12549
    UPLIFTJEE
    Keymaster

    Answer

    C

  • Author
    Posts
Log In Register Lost Password
Viewing 3 posts - 1 through 3 (of 3 total)
  • You must be logged in to reply to this topic.