Please answer 87th question


$$\begin{gathered} \mathbf{87}\mathbf{.}\mathbf{ }\mathbf{ } The locus of the mid points of the portion of the \tan gents to the ellipse \frac{x^2}{16} + \frac{y^2}{9}=1 intercepted between the axes is \\ \left(1\right) \frac{16}{x^2} + \frac{9}{y^2} = 4 \left(2\right) \frac{x^2}{16} + \frac{y^2}{9} = 4 \\ \left(3\right) \frac{x^2}{16} + \frac{y^2}{9} = 16 \left(4\right) \frac{x^2}{16} + \frac{y^2}{9} = 25 \end{gathered}$$