Question

Assertion A: The least length of the focal chord of $y^2=4ax$ is $4a$.
Reason R: Length of the focal chord of $y^2=4ax$ which makes an angle $\theta$ with its axis is $4a{\text{cosec}}^2\theta$.

• A. Both A and R are true and R is the correct explanation of A.
• B. Both A and R are true and R is not the correct explanation of A.
• C. A is true but R is false.
• D. A is false but R is true.

Answer 1

The correct option is A Both A and R are true and R is the correct explanation of A.

Lead Length of a focal chord in the length of the lotus ractem.
$\tan \theta =m$

$(y-0)=m(x-a)$

$m^2(x-a{)}^2=40x$

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

$\Rightarrow L=\sqrt{(x_1-x_2{)}^2+m^2(x_1-x_2{)}^2}$

$\Rightarrow L=\sqrt{\left(\right(x_1+x_2{)}^2-4x_1{x}_2\left)\right(1+m^2)}$ ...(1)

$\Rightarrow m^2-x^2-2ax(2+m^2)+a^2{m}^2=0$

$\Rightarrow x_1+x_2=\frac{2a(2+m^2)}{m^2}$
$\Rightarrow \left(x_1{x}_2\right)=a^2$

On subsetituting in $e{q}^n\left(1\right)$, we get $l=4a\left(\frac{m^2+1}{m^2}\right)$
$\therefore l=4a{\text{cosec}}^2\theta$

Hence, minimum length of focal child will be $4a$.