AB is a chord of the parabola y2=4ax with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is

Question

AB is a chord of the parabola y2=4ax with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is (A) 2 (B)  2a  (C)  4a  (D)  8a

Tardigrade Admin · Accepted Answer

In Δ ABC, we have tan θ -(y/x) ...(i)In Δ BCD, we have tan (90° -θ )=(y/CD) ⇒ CD=y tan θ =(y2/x) [using Eq. (i)] ⇒ CD=(4ax/x)=4a

Q. AB is a chord of the parabola $y^{2} = 4 a x$ with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is

2

$2 a$

$4 a$

$8 a$

In $Δ A BC,$ we have $tan θ - \frac{y}{x}$ ...(i)
In $Δ BC D,$ we have $tan (90^{\circ} - θ) = \frac{y}{C D}$ $\Rightarrow$ $C D = y tan θ = \frac{y ^{2}}{x}$ [using Eq. (i)] $\Rightarrow$ $C D = \frac{4 a x}{x} = 4 a$

Q. AB is a chord of the parabola y2=4ax with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is

2

2a

4a

8a

In ΔABC, we have tanθ−xy​ ...(i)In ΔBCD, we have tan(90∘−θ)=CDy​ ⇒ CD=ytanθ=xy2​ [using Eq. (i)] ⇒ CD=x4ax​=4a

Q. AB is a chord of the parabola $y^{2} = 4 a x$ with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is

$2 a$

$4 a$

$8 a$

In $Δ A BC,$ we have $tan θ - \frac{y}{x}$ ...(i)
In $Δ BC D,$ we have $tan (90^{\circ} - θ) = \frac{y}{C D}$ $\Rightarrow$ $C D = y tan θ = \frac{y ^{2}}{x}$ [using Eq. (i)] $\Rightarrow$ $C D = \frac{4 a x}{x} = 4 a$