Question

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

• A. $4acose{c}^2\theta$
• B. $4ase{c}^2\theta$
• C. $acose{c}^2\theta$
• D. none of these

Answer 1

Answer: (A)

$4acose{c}^2\theta$

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 $=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$=\sqrt{(\frac{y_1^2}{4a}-\frac{y_2^2}{4a}{)}^2+(y_1-y_2{)}^2}$

$=(y_1-y_2)\sqrt{(\frac{y_1+y_2}{4a}{)}^2+1}$

Equation of line is

$y=tan\theta (x-a)$

$⟹x=ycot\theta +a$

Substituting this in the equation of parabola, we get

$y^2=4ax=4a(ycot\theta +a)$

$\therefore y^2-4acot\theta y-4a^2=0$

$⟹y_1+y_2=4acot\theta$

$⟹y_1-y_2=\sqrt{(y_1+y_2{)}^2-4y_1{y}_2}$

$=\sqrt{16a^{2}co{t}^2\theta +16a^2}$

$=\sqrt{16a^{2}cose{c}^2\theta }$

$=4acosec\theta$

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

$=4acosec\theta \sqrt{(\frac{4acot\theta }{4a}{)}^2+1}$

$=4acose{c}^2\theta$

So, the answer is option (A).