Question

If a focal chord $y^2=4ax$ makes an angle $\alpha ,\alpha ϵ(0,\frac{\pi }{4}]$ with the positive direction of x- axis, then minimum length of this focal chord is

• A. $6a$
• B. $2a$
• C. $8a$
• D. None of the above

Answer 1

The correct option is C $8a$

If focal chord $y^2=4ax$ makes an angle $\alpha ,\alpha \in (0,\frac{\pi }{4})$

With the positive direction of $x-$ axis, then minimum length of this focal chord is

Minimum length is when it has maximum angle i.e. $\alpha =\frac{\pi }{4}$

$m=1$

Equation of focal chord is

$y=x-a{y}^2=4ax$

Solve both equations

$x^2+a^2-2ax=4ax$

$x^2-6ax+a^2=0$

$x=\frac{6a\pm \sqrt{36a^2-4a^2}}{2}$

$=3a\pm 2\sqrt{2}$

$=(3\pm 2\sqrt{2})a$

$y=(3\pm 2\sqrt{2})a-a=(2\pm 2\sqrt{2})a$

$A=((3+2\sqrt{2})a,(2+2\sqrt{2})a)$

$B=((3-2\sqrt{2})a,(2-2\sqrt{2})a)$

$AB=a\sqrt{{(3+2\sqrt{2}-3+2\sqrt{2})}^2+{(2+2\sqrt{2}-2+2\sqrt{2})}^2}$

$=a\sqrt{{\left(4\sqrt{2}\right)}^2+{\left(4\sqrt{2}\right)}^2}$

$=\sqrt{64}a$

$=8a$ Units