Ans.

(i)

Sol. As y = mx – 2am – am 3 (making the equation homogenous with curve

y

2 = 4ax)

⇒ y 2 =

⇒ y 2 (m 3 + 2m) – 4mx 2 + 4xy = 0 ......

Now angle between the lines represented by is 90º

Coefficient of x 2 + coefficient of y 2 = 0

⇒ m 3 + 2m – 4m = 0, m 3 – 2m = 0

⇒ m = 0, m = ± (but m ≠ 0)

∴ m = , –

(ii)

Sol. Let PCP ' and QCQ ' be a pair of conjugate diameters of an ellipse = 1.

Then P (acos φ 1 , bsin φ 1 ), as centre of ellipse is C(0, 0) so m 1 = slope of CP

= = tan φ 1 and m 2 = slope CQ = tan φ 2

m 1 m 2 = tan φ 1 tan φ 2 ..........

Since the diameters PCP ' and QCQ ' are conjugate diameters

∴ m 1 m 2 = – ............

from & we have tan φ 1 tan φ 2 = – 1

sin φ 1 sin φ 2 + cos φ 1 cos φ 2 = 0

⇒ cos ( φ 1 – φ 2 ) = 0

⇒ φ 1 – φ 2 = ±

(iii)

Sol. Let ( λ , m) be mid points of P (acos φ , bsin φ ) and Q (– asin φ , bcos φ ) is given by-

λ = (cos φ – sin φ ) & m = (sin φ + cos φ )

= cos φ – sin φ and = sin φ + cos φ ....(*)

⇒ + = 2

(on squaring & adding the retations of *)

⇒ + =

So locus of mid of PQ

+ = is the required locus