Assuming P lies on H
Origin: O(0,0)
Point P: P(x_1, y_1)
H: x^2/a^2 + y^2/b^2 = 1
OP is perpendicular to Tangent @ P
--> ie, (m_op).(m_tangent@P) = -1 <-- the condition required
m_op = (y_1-0)/(x_1-0) = y_1/x_1
m_tangent@P:
Differentiate H wrtx: (2x/a^2) + (2y/b^2)*dy/dx = 0 (implicit differentiation and chain rule)
Rearrange to find dy/dx: dy/dx = -b^2.x/a^2.y
Substitute P(x_1, y_1) to find gradient at P: m_tangent@P = -b^2.x_1/a^2.y_1
Now, continue with: (m_op).(m_tangent@P) = -1
(y_1/x_1)*(-b^2.x_1/a^2.y_1) = -1
Simplify --> b^2 = a^2
Consider the given Locus of P: (x^2+y^2)^2 = a^2.x^2 + b^2.y^2
Replacing b^2 with a^2 gives us: (x^2+y^2)^2 = a^2.x^2 + a^2.y^2
= a^2.(x^2+y^2)
Cancelling (x^2+y^2) on both sides: x^2 + y^2 = a^2 (x^2 +y^2 cannot = 0)
Which represents a circle with centre (0,0) and radius a units, or, if we replaced a^2 with b^2, a radius of b units
Point P lying on H now becomes H: x^2/a^2 + y^2/a^2 = 1 --> x^2 + y^2 = a^2
i.e. P lies on a circle with radius a units or radius b units
