Length of normal. chord...

Question

What is the length of normal chord which subtends right angle at the vertex of parabola $y^2=4x$. $$My Try$$ let the equation of normal be $y=mx-am^3-2am$ Now I assumed slope of this line as $45$ (which is probably wrong). So I got $y=x-3$ thus solving for $y^2=4x$ and the calculated equation I got $x=9,y=6$ ignoring the smaller root. As answers given are more in magnitude. So I got length from focus $1,0$ as $10$ but it is wrong. Thanks!

Answer 1

"Subtends right angle"? Do you mean a line segment AB such that A and B lie on the parabola and angle AOB is a right angle? IF the A and B are $(x, 2\sqrt{x})$ and $(x,-2\sqrt{x})$ Then the length of the line segment is, of course, just the difference in y values, $4\sqrt{x}$. That would be the case when the two sides of the angle make 45 degree angles with the x-axis.

But that is not the only possible case. If, say, the upper side makes angle $\theta$ with the x-axis, it can be written $y= tan(\theta)x$ and crosses the parabla where $tan(\theta)x= 2\sqrt{x}$ so $\sqrt{x}= 2cot(\theta)$, $x= 4 cot^2(\theta)$, $y= 2\sqrt{x}= 2(2 cot(\theta)= 4 cot(\theta)$. The other line, which is perpendicular to the first, so has slope $-\frac{1}{tan(\theta)}= -cot(\theta)$, has equation $y= -cot(\theta)x$ crosses the parabola where $-cot(\theta)x= -2\sqrt{x}$ so $\sqrt{x}= -2tan(\theta)$, $x= 4 tan^2(\theta)$, $y= -4 tan(\theta)$. Can you find the distance between points $A= (4 cot^2(\theta), 4 cot(\theta))$ and $B= (4 tan(\theta), -4\tan(\theta))$?

Answer 2

Let the normal chord be $AB$ where $A$ and $B$ lie on the parabola $y^2 = 4x$. Points on this parabola will be of the form $(t^2, 2t)$ where $t$ is a parameter.

Let $A$ have the parameter $t$. Then $B$'s parameter will be given by $t' = -t - \frac{2}{t}$ as $AB$ is a normal chord.

Now, as $AB$ subtends a right angle at the vertex $O(0,0)$, the product of slopes of $OA$ and $OB$ is equal to $-1$

$$ m_{OA} \cdot m_{OB} = -1 $$ $$ \implies \frac{2t}{t^2} \cdot \frac{2t'}{t'^2} = -1$$ $$ \implies tt' = -4$$ $$ \implies t^2 + 2 = 4$$ $$ \implies t = \pm\sqrt2$$ Consider any one of the two roots and find the corresponding $t'$ from the previous relation. Now, you have the coordinates of both $A$ and $B$. The length of $AB$ can be found out using distance formula.