Question

Normal chord of a parabola y^2=4ax at point whose ordinate is equal to abscissa,then angle subtended by normal chord to at focus is

Answer 1

Answer:

90°

Step-by-step explanation:

Hi,

Point whose ordinate equal to abscissa will be such that x = y,

but (x, y) lies on y² = 4ax

⇒ x² = 4ax

⇒ x = 0 or x = 4a

Since at (0,0) we cannot draw a normal chord, hence point  at which normal chord is drawn is P (4a, 4a) .

Focus of the given parabola y² = 4ax is S(a, 0).

Equation of the normal chord at ( 4a, 4a) will be

y - 4a/x - 4a = -2

⇒ 2x + y = 12a is the equation of normal.

To find the points of intersection of normal with the parabola , we need to

solve the equation  (12a - 2x)² = 4ax

⇒ (6a - x)² = ax

⇒ 36a² - 12ax + x² = ax

⇒ x² - 13ax + 36a² = 0

⇒(x - 4a)(x-9a) = 0

⇒ x = 4a or 9a

P (4a, 4a) and Q (9a, -6a) are the end points of the normal chord.

Slope of line PS = (4a - 0)/(4a - a) = 4/3

Slope of line QS = (-6a - 0)/(9a - a) = -3/4

We can observe that product of (Slope of PS)*(Slope of QS) = -1

⇒ ∠PSQ = 90°.

Thus, normal chord at the point whose ordinate is equal to abscissa subtends an angle of 90° at the focus.

Hope it helps !