  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  B
4a sec2θ  C
a cosec2θ  D
none of these  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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