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Question

The length of a focal chord of the parabola $$y^2 = 4ax$$ making an angle with the axis of the parabola is


A
4a cosec2θ
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B
4a sec2θ
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C
a cosec2θ
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D
none of these
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Solution

The correct option is C $$4a \ cosec^2\theta$$
The axis of parabola $$y^2=4ax$$ is X-axis and focus is $$(a,0)$$
Equation of line passing through $$(a,0)$$ and making and angle $$\theta$$ with X-axis is $$y=tan\theta(x-a)$$

Let this line intersect the parabola at $$(x_1,y_1)$$ and $$(x_2,y_2)$$
$$\therefore$$ Length of focal cord $$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
$$=\sqrt{(\dfrac{y_1^2}{4a}-\dfrac{y_2^2}{4a})^2+(y_1-y_2)^2}$$
$$=(y_1-y_2)\sqrt{(\dfrac{y_1+y_2}{4a})^2+1}$$

Equation of line is
$$y=tan\theta(x-a)$$
$$\implies x=ycot\theta+a$$
Substituting this in the equation of parabola, we get
$$y^2=4ax=4a(ycot\theta+a)$$

$$\therefore y^2-4acot\theta y-4a^2=0$$
$$\implies y_1+y_2=4acot\theta$$     
$$\implies y_1-y_2=\sqrt{(y_1+y_2)^2-4y_1y_2}$$
                        $$=\sqrt{16a^2cot^2\theta+16a^2}$$
                        $$=\sqrt{16a^2cosec^2\theta}$$
                        $$=4acosec\theta$$

Length of focal cord $$=(y_1-y_2)\sqrt{(\dfrac{y_1+y_2}{4a})^2+1}$$
$$=4acosec\theta\sqrt{(\dfrac{4acot\theta}{4a})^2+1}$$
$$=4acosec^2\theta$$

So, the answer is option (A).



Mathematics

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