Find the locus of the point of intersection of the tangents of the ellipse $b^2x^2+a^2y^2=a^2b^2$ at the points $\phi$ and $\frac{\pi}{2}+\phi$. - "a" and "b" are constants.
- "x" and "y" are variables.
- pi=22/7
- ø= an angle

I try this question at tangent method. But perhaps I do some mistake.
I take the tangents when intersect make angle $\phi$ and $\dfrac{\pi}{2}+\phi$ and then calculate for finding locus. But I failed at that way.

Let $(x',y')$ be the point of intersection, then the equation of the polar line (i.e. the chord in this case) is
$$\frac{x'x}{a^2}+\frac{y'y}{b^2}=1 \tag{1}$$
End points of the chord:
$$(a\cos \phi,b\sin \phi) \: \text{ and } \: (-a\sin \phi,b\cos \phi)$$
Hence the equation of the chord is
\begin{align*}
  \frac{y-b\sin \phi}{x-a\cos \phi} &=
  \frac{b\sin \phi-b\cos \phi}{a\cos \phi+a\sin \phi} \\
  (y-b\sin \theta)(a\cos \phi+a\sin \phi) &=
  (x-a\cos \phi)(b\sin \phi-b\cos \phi) \\
  b(\cos \phi-\sin \phi)x+a(\cos \phi+\sin \phi)y &=
  ab(\cos^2 \phi+\sin^2 \phi) \\
  \frac{(\cos \phi-\sin \phi)x}{a}+\frac{(\cos \phi+\sin \phi)y}{b} &= 1 \tag{2}
\end{align*}
Comparing $(1)$ and $(2)$,
$$\begin{pmatrix} x' \\ y' \end{pmatrix} =
\begin{pmatrix}
  a(\cos \phi-\sin \phi) \\
  b(\cos \phi+\sin \phi)
\end{pmatrix}$$
which lies on the ellipse
$$\frac{x^2}{2a^2}+\frac{y^2}{2b^2}=1$$
