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AB is a chord...

Question

$A B$ is a chord of the parabola $y^{2} = 4 a x$ with vertex at $A$ is drawn perpendicular to $A B$ meeting the axes at $C$ , the projection of $B C$ on the axis of the parabola is

a

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2 a

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4 a

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8 a

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Solution

The correct option is C $4 a$
Given equation of parabola is $y^{2} = 4 a x$
In $△ A B D$ , we have
$tan θ = \frac{y}{x}$ ....(i)
In $△ B C D$ , we have
$tan (90^{o} - θ) = \frac{y}{C D}$
$\Rightarrow C D = y tan θ = \frac{y^{2}}{x}$
$\Rightarrow C D = \frac{4 a x}{x} = 4 a$

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

The correct option is C 4aGiven equation of parabola is y2=4axIn △ABD, we havetanθ=yx ....(i)In △BCD, we havetan(90o−θ)=yCD⇒CD=ytanθ=y2x⇒CD=4axx=4a

$A B$ is a chord of the parabola $y^{2} = 4 a x$ with vertex at $A$ is drawn perpendicular to $A B$ meeting the axes at $C$ , the projection of $B C$ on the axis of the parabola is

The correct option is C $4 a$
Given equation of parabola is $y^{2} = 4 a x$
In $△ A B D$ , we have
$tan θ = \frac{y}{x}$ ....(i)
In $△ B C D$ , we have
$tan (90^{o} - θ) = \frac{y}{C D}$
$\Rightarrow C D = y tan θ = \frac{y^{2}}{x}$
$\Rightarrow C D = \frac{4 a x}{x} = 4 a$