Question

The length of focal chord which makes an angle $\alpha$ with the axis. of the parabola $y^2=4ax$ is

• A. $2aco{t}^2\alpha$
• B. $4acose{c}^2\alpha$
• C. $4acot\alpha$
• D. None of these

Answer 1

Answer: (B)

Let $p(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$ be the end points of the focal chord PQ of the parabola $y^2=4ax$ and which makes an angle $\alpha$ with the axis of parabola.
Now, $PQ=\sqrt{{(a{t}_2^2-a{t}_1^2)}^2+{(2a{t}_2-2a{t}_1)}^2}$ $=a\sqrt{{(t_2^2-t_1^2)}^2+4{(t_2-t_1)}^2}$ $=a\sqrt{{(t_2+t_1)}^2{(t_2-t_1)}^2+4{(t_2-t_1)}^2}$ $=a(t_2-t_1)\sqrt{t_2^2+t_1^2+2t_1{t}_2+4}$ $=a(t_2-t_1)\sqrt{t_2^2+t_1^2-2+4}$ $(\because t_1{t}_2=-1)$ $=a(t_2-t_1)\sqrt{t_2^2+t_1^2+2}$ $=a(t_2-t_1)\sqrt{(t_2^2+t_1^2-2t_1{t}_2)}$ $(\because t_1{t}_2=-1)$ $PQ=a(t_2-t_1)\sqrt{{(t_2-t_1)}^2}$ $=a{(t_2-t_1)}^2$ and slope of $PQ=\frac{2a{t}_2-2a{t}_1}{a{t}_2^2-a{t}_1^2}$ $\tan \alpha =\frac{2a(t_2-t_1)}{a(t_2-t_1)(t_2-t_1)}$ $\Rightarrow$ $\tan \alpha =\frac{2}{t_1+t_2}$ $\Rightarrow$ $2\cot \alpha =t_1+t_2$ Again, we have Eq. (i) $PQ=a{(t_2-t_1)}^2$ $=a[{(t_2+t_1)}^2-4t_1{t}_2]$ $=a[{(2\cot \alpha )}^2+4]$ $[\because t_1+t_2=2\cot \alpha and{t}_1{t}_2=-1]$ $=4a(1+{\cot }^2\alpha )$ $PQ=4a\cos e{c}^2\alpha$