Q.395. Length of normal at an extremity of latus rectum intercepted by parabola

Question 395.
If the length of normal at an extremity of latus rectum of parabola y^2 = 4ax , a > 0 intercepted by parabola is 4 units,
find a.

Answer 395.
Extremity of latus rectum is (a, 2a)
y^2 = 4ax
=> 2y dy/dx = 4a
=> dy/dx = 2a/y
=> slope of normal = dx/dy = - y/2a
=> slope of normal at (a, 2a) = - 1
=> eqn. of normal at (a, 2a) is
y - 2a = - 1 (x - a)
=> y = - x + 3a
Solving with y^2 = 4ax,
(-x + 3a)^2 = 4ax
=> x^2 - 10ax + 9a^2 = 0
=> (x - 9a) (x - a) = 0
=> x = a or 9a
x = 9a => y = - 9a + 3a = - 6a
=> the other point of intersection of normal with parabola is
(9a, - 6a)
=> 4^2 = (9a - a)^2 + (- 6a - 2a)^2
=> 16 = 128 a^2
=> a = 1/(2√2).

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