A rectangle OABC is drawn so that O is the vertex of parabola y=x^2 and OA and OB are two chords drawn on the parabola. What is the locus of point C?

(A). $y = {x}^{2} + 2$ (B). $y = - {x}^{2} + 4$ (C). $y = 2 {x}^{2}$ (D). $y = - 2 {x}^{2} + 4$

Jan 13, 2018

(A) $y = {x}^{2} + 2$

Explanation:

The parabola $y = {x}^{2}$ is symmetric w.r.t. $y$-axis. Further as $O A B C$ is a rectangle, its each angle is ${90}^{\circ}$ i.e. $m \angle A O B = {90}^{\circ}$.

As the sides $A O$ and $B O$ pass through origin at $O \left(0 , 0\right)$, their equation will be will be of type $y = m x$ and $y = - \frac{1}{m} x$ or $x + m y = 0$.

Now intersection of $y = m x$ with parabola will be given by $m x = {x}^{2}$ i.e. $x = m$ and the corresponding point is $\left(m , {m}^{2}\right)$. Similarly intersection of $x + m y = 0$ gives $x + m {x}^{2} = 0$ i.e. $x = - \frac{1}{m}$ and the corresponding point is $\left(- \frac{1}{m} , \frac{1}{m} ^ 2\right)$.

As $A \left(m , {m}^{2}\right)$ and $B \left(- \frac{1}{m} , \frac{1}{m} ^ 2\right)$, we must have $C \left(m - \frac{1}{m} , {m}^{2} + \frac{1}{m} ^ 2\right)$ and as

${\left(m - \frac{1}{m}\right)}^{2} = {m}^{2} + \frac{1}{m} ^ 2 - 2$, desired equation is $y = {x}^{2} + 2$ and answer is (A).

Below is shown the graph and rectangle relating to $m = 2$.

graph{(y-x^2)(2y+x)(y-2x)(8x-4y+5)(2x+4y-20)(y-x^2-2)=0 [-4.84, 5.16, -0.38, 4.62]}

Jan 14, 2018

I think that name of the rectangle should be $O A C B$ not $O A B C$

Let the coordinates of

A $\to \left({t}_{1} , {t}_{1}^{2}\right)$
B $\to \left({t}_{2} , {t}_{2}^{2}\right)$
C $\to \left(h , k\right)$
O $\to \left(0 , 0\right)$

Gradient of OA $= {t}_{1}^{2} / {t}_{1} = {t}_{1}$
Gradient of OB $= {t}_{2}^{2} / {t}_{2} = {t}_{2}$

As OA and OB are adjacent sides of the rectangle then product of their gradients should be $- 1$
Hence t_1*t_2=-1... .(1)

Now the diagonals ºC and $A B$ should bisect each other.
So coordinates of mid point of $O C \to \left(\frac{h}{2} , \frac{k}{2}\right)$

And coordinates of mid point of $A B \to \left(\frac{{t}_{1} + {t}_{2}}{2} , \frac{{t}_{1}^{2} + {t}_{2}^{2}}{2}\right)$

Hence $h = {t}_{1} + {t}_{2} \mathmr{and} k = \left({t}_{1}^{2} + {t}_{2}^{2}\right)$

So $k = \left({t}_{1}^{2} + {t}_{2}^{2}\right)$
$\implies k = {\left({t}_{1} + {t}_{2}\right)}^{2} - 2 {t}_{1} {t}_{2}$
$\implies k = {h}^{2} - 2 \left(- 1\right)$
$\implies k = {h}^{2} + 2$

Converting $\left(h , k\right) \to \left(x , y\right)$ we get the locus of $C$ as follows

color(green)(y=x^2+2, which is option $\left(A\right)$