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!
[Math] Length of normal. chord…
analytic geometryconic sections
Related Solutions
Let $k := \overleftrightarrow{PQ}$ be a line containing focal chord $\overline{PQ}$. Let lines $\ell_P$ and $\ell_Q$, through $P$ and $Q$, respectively, be parallel to the parabola's axis.
By the Reflection Property of Parabolas, the normal and tangent at $P$ bisect angles made by $k$ and $\ell_P$; likewise, at $Q$. In particular, the tangent at $P$ (call it $t$) and the normal at $Q$ (call it $n$) bisect alternate interior angles formed by parallel lines $\ell_P$ and $\ell_Q$ cut by transversal $k$. These angles are congruent, so their half-angles are congruent. A pair of such half-angles comprise the alternate interior angles of $t$ and $n$ cut by $k$, whence $t \parallel n$: the tangent at $P$ is parallel to the normal at $Q$.
Proof without words
To find the vertex of that parabola, you can make use of a nice general result:
given three tangents of a parabola, if $AB$ is that part of a tangent which is comprised between the other two tangents (see diagram), then the projection of $AB$ onto a line perpendicular to the axis has a fixed length, independent of the position of $AB$.
In our particular case, if $V$ (vertex) and $D$ are the tangency points of the "outer" tangents, $C$ is their intersection point, and $B'$, $D'$ are the projections of $B$, $D$ on tangent $VC$ (which is perpendicular to the axis), we obtain from the above result:
$$AB'=CD'=VC.$$
As points $A$, $B$, $C$, $B'$ can be readily found from the data, one can also easily find vertex $V$ and tangency point $D$.
Best Answer
"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 $B= (4 tan(\theta), -4\tan(\theta))$?