Official solution
Given, 4x2 - 9y2 = 36
After differentiating w.r.t.x, we get
4.2x - 9.2.y.\[{{dy} \over {dx}}\] = 0
\[ \Rightarrow \] Slope of tangent = \[{{dy} \over {dx}}\] = \[{{4x} \over {9y}}\]
So, slope of normal = \[{{ - 9y} \over {4x}}\]
Now, equation of normal at point (x0, y0) is given by
y - y0 = \[{{ - 9{y_0}} \over {4{x_0}}}\] (x - x0)
As normal intersects X axis at A, Then
A = \[\left( {{{13{x_0}} \over 9},0} \right)\] and B \[ \equiv \] \[\left( {0,{{13{y_0}} \over 4}} \right)\]
As OABP is parallelogram
\[ \therefore \] midpoint of OB \[ \equiv \] \[\left( {0,{{13{y_0}} \over 8}} \right)\] \[ \equiv \] Midpoint of AP
So, P(x, y) \[ \equiv \] \[\left( {{{ - 13{x_0}} \over 9},{{13{y_0}} \over 4}} \right)\] ...(i)
\[ \because \] (x0, y0) lies on hyperbola, therefore
4(x0)2 - 9(y0)2 = 36
From equation (i) : x0 = \[{{ - 9x} \over {13}}\] and y0 = \[{{4y} \over {13}}\]
From equation (ii), we get
9x2 - 4y2 = 169
Hence, locus of point P is : 9x2 - 4y2 = 169