Question

A particle is placed on the parabola $y=x^2-x-6$ at a point $P$ whose ordinate is $6$. It is allowed to roll along the parabola until it reaches the nearest point $Q$ whose ordinate is $-6$. The horizontal distance travelled by the particle (the numerical value of the difference in the abscissas of $P$ and $Q$) is:

• A. $5$
• B. $4$
• C. $3$
• D. $2$
• E. $1$

Answer 1

The correct option is C $3$

Given Parabola: $y=x^2-x-6...............\left(1\right)$

Let the point $P$ be $\left(\right(x_1,6)$and$Q$be$(x_2,-6)$

Then both lies on parabola at,

$P,$ equation $\left(1\right)=>6=x_1^2-x_1-6$

$=>x_1^2-x_1-12=0$

$=>x_1=-3,4$

Lets's consider the value of $x_1$ for left from the axis of parabola, So,

$P:(-3,6)$

Now at $Q$, equation $\left(i\right)$, $=>-6=x_2^2-x_2-6$

$=>x_2^2-x_2=0$

$=>x_2=0,1$

Let consider the lower abssica, So, $Q:(0,-6)$

now horizontal displacement $=|x_2-x_1|$

$=0-(-3)$

$=3.$