Question

The length of the chord of the parabola $x^2=4y$ passing through the vertex and having slope $cot\alpha$ is

• A. $4\cos \alpha .cose{c}^2\alpha$
• B. $4a\tan \alpha \sec \alpha$
• C. $4\sin \alpha .{\sec }^2\alpha$
• D. none of these

Answer 1

The correct option is B $4a\tan \alpha \sec \alpha$

Equation of parabola $x^2=4ay$

Vertex of parabola $\left(0\right)=(0,0)$

Equation of chord passing through vertex $y-y_1=m(x-x_1)$

Here, $x_1=0$ and $y_1=0$

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

$\therefore y=mx$

Now, slope$=m=\tan \alpha$

Let the chord intersect the parabola at $P(h,k)$

Since, point P lies on chord

$k=\left(\tan \alpha \right)h$ ………..$\left(1\right)$

Point P lies on parabola

$h^2=4ak$ ………$\left(2\right)$

Substituting equation $\left(1\right)$ in $\left(2\right)$

$\therefore h^2=4a(\tan \alpha \times h)$

$\therefore h^2(h-4a\tan \alpha )=0$

$\therefore h=0$

$h=4a\tan \alpha$

Substituting $h=4a\tan \alpha$ in equation $\left(2\right)$, we get

$\therefore (4a{)}^2{\tan }^2\alpha =4ak$

$\therefore k=4a{\tan }^2\alpha$

$\therefore p=(h,k)=(4a\tan \alpha ,4a{\tan }^2\alpha )$

Distance $OP=\sqrt{(4a\tan \alpha -0{)}^2+(4a{\tan }^2\alpha -0)}$

$=\sqrt{16a^2{\tan }^2\alpha +16a^2{\tan }^4\alpha }$

$=4a\tan \alpha \sqrt{1+{\tan }^2\alpha }$

$=4a\tan \alpha \sqrt{{\sec }^2\alpha }$

$=4a\tan \alpha \sec \alpha$.