Question

Prove that the normal chord at the point other than origin whose ordinate is equal to its abscissa subtends a right angle at the focus. 

Answer 1

Let the equation of the parabola be y² = 4ax and P(at² , 2at) be any point …(1) On the parabola for which the abscissa is equal to the ordinate. 

 i.e. at² = 2at ⇒ t = 0 or t = 2. But t ≠ 0.

Hence the point (4a, 4a) at which the normal is

 y + 2x = 2a(2) + a(2)³ 

y = (12a – 2x) ……….… (2) 

Substituting the value of 

 y = (12a – 2x) in (1) we get 

(12a – 2x)² = 4ax 

x² – 13ax + 36a² = (x – 4a)(x – 9a) = 0 

⇒ x = 4a, 9a 

Corresponding values of y are 4a and –6a. 

Hence the other points of intersection of that normal at P(4a, 4a) to the given parabola is Q(9a, –6a), we have S(a, 0). 

Slope of the SP=

Slope of the SQ=

Clearly m₁m₂ = –1, so that SP⊥ SQ .