A normal to the hyperbola 4x² - 9y² = 36 meets the coordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is

given hyperbola is: 4x² - 9y² = 36
let (x₀, y₀) be point of contact of normal on the hyperbola
Finding slope of normal at that point:
Differntiating hyperbola equation we get;
4 × 2 × x - 9 × 2 × y(dy/dx) = 0
⇒ dy/dx = 4x/9y = slope of tangent
∴slope of normal = -9y/4x
equation of normal at (x₀, y₀) is:
y - y₀ = -9y₀/4x₀ (x - x₀)
line intersects X axis at A when y = 0
∴A = (13x₀/9, 0)
similarly B = (0, 13y₀/4)
given OABP forms a parallelogram ⇒ diagonals bisect each other
(midpoint of diagonals are same)
midpoint of OB = (0, 13y₀/8) = midpoint of AP
Let P = (x, y) ∴ midpoint of AP = ((13x₀/9 + x)/2, y/2)
∴P(x, y) = (13x₀/9, 13y₀/4) -------(1)
As (x₀, y₀) lie on hyperbola, it should satisfy its equation:
4(x₀)² - 9(y₀)² = 36
from equation (1): x₀ = 9x/13 and y₀ = 4y/13
substituting in hyperbola equation , we get:
4(9x/13)² - 9(4y/13)² = 36
324x²/169 - 144y²/169 = 36
9x² - 4y² = 169
∴locus of point P is hyperbola whose equation is:
9x² - 4y² = 169
hence correct option is D.