If the normal drawn from the point on the axis of the parabola y 2= 8ax whose distance from the focus is 8 a and which is not parallel to either axis, makes an angle. θ with the axis of x, then θ is equal to

Question

If the normal drawn from the point on the axis of the parabola y 2= 8ax whose distance from the focus is 8 a and which is not parallel to either axis, makes an angle. θ with the axis of x, then θ is equal to (A) (π/6) (B) (π/4) (C) (π/3) (D) (2 π/3)

Tardigrade Admin · Accepted Answer

The focus of the parabola y2=8 ax is (2 a, 0). So, the coordinates of the point on the axis of the parabola at a distance 8 a from the focus is (10 a , 0). Equation of a normal to the parabola y 2=8 ax is y=m x-4 a m-2 a m3 Since it passes through (10 a , 0), ∴ 0=10 am -4 am -2 am 3 ⇒ 2 am (3- m 2)=0 ⇒ m 2=3(∵ m ≠ 0) ⇒ m=± √3= tan (± (π/3))

Q. If the normal drawn from the point on the axis of the parabola $y^{2} = 8 a x$ whose distance from the focus is $8 a$ and which is not parallel to either axis, makes an angle. $θ$ with the axis of $x$ , then $θ$ is equal to

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{2 π}{3}$

Q. If the normal drawn from the point on the axis of the parabola y2=8ax whose distance from the focus is 8a and which is not parallel to either axis, makes an angle. θ with the axis of x, then θ is equal to

6π​

4π​

3π​

32π​

Q. If the normal drawn from the point on the axis of the parabola $y^{2} = 8 a x$ whose distance from the focus is $8 a$ and which is not parallel to either axis, makes an angle. $θ$ with the axis of $x$ , then $θ$ is equal to

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{2 π}{3}$