Sol. The equation of any tangent to + = 1 is

y = mx ± ... (i)

Let P(h, k) be the foot of the perpendicular drawn from the origin O(0, 0) to (i).

Since (h, k) lies on (i). Therefore,

k = mh ± ... (ii)

Slope of OP =

Since OP is perpendicular to (i). Therefore,

× m = –1 ⇒ m =

Putting the value of m in (ii), we get

k = – ±

⇒ (k 2 + h 2 ) 2 = a 2 h 2 + b 2 k 2 Thus the locus of (h,k) is, (x

2 + y 2 ) 2 = a 2 x 2 + b 2 y 2 ... (iii)

We have to find the differential equation satisfied by (iii).

Differentiating (iii) w.r.t.x, we get

2(x 2 + y 2 ) = 2a 2 x + 2b 2 y

⇒ 2(x 2 + y 2 ) = a 2 x + b 2 y ... (iv)

Differentiating (iv) with respect to x, we get

4 + 2 (x 2 + y 2 )

= a 2 + b 2 ... (v)

Multiplying (iv) by x and subtracting from (iii), we get

(x 2 + y 2 )

⇒ (x 2 + y 2 ) = b 2 y ... (vi)

Multiplying (v) by x and subtracting from (iv), we get

2(x 2 + y 2 )

– 4 = b 2

⇒ 2(x 2 + y 2 ) – 4

= b 2 ... (vii)

Eliminating b 2 from (vi) and (vii), we obtain the required differential equation as

=