Question

If normal at $P(18,12)$to the parabola ${y^2} = 8x$ cuts it again at Q , Show that $9PQ = 8\sqrt {109}$.

Answer 1

Hint: Here, we find the value of $a$ by comparing the general equation of a parabola ${y^2} = 4ax$ with the given equation of a parabola. Further, we use this value of $a$ to find the coordinates of points on the parabola and then using the formula for length of a line joining two points $(x,y)$ and $(a,b)$ $L = \sqrt {{{(x - a)}^2} + {{(y - b)}^2}}$ we find the distance between two points.

Complete step by step solution:
Given a parabola${y^2} = 8x$ , firstly find the points of the parabola
When $x = 0$ , ${y^2} = 8 \times 0 = 0$ , so, $y = 0$
When $x = 2$ , ${y^2} = 8 \times 2 = 16$ , so, $y = \sqrt {16} = \pm 4$
When $x = 8$ , ${y^2} = 8 \times 8 = 64$ , so, $y = \sqrt {64} = \pm 8$
Therefore some coordinates of the parabola ${y^2} = 8x$ are $(0,0),(2, - 4),(2,4),(8, - 8),(8,8)$
Now we plot the graph of the parabola ${y^2} = 8x$

We can clearly see the parabola ${y^2} = 8x$ lies in Quadrant $1$ and quadrant $4$.
On comparing the parabola ${y^2} = 8x$ to ${y^2} = 4ax$ (standard form of parabola), we get
$8x = 4ax$
i.e. $a = \dfrac{{8x}}{{4x}} = 2$ $...(i)$
As we know any coordinates of normal on the parabola can be written as $(a{t^2},2at)$ where $t$ is a point on the normal.
Therefore substituting the value of $a = 2$ from equation $(i)$
Coordinates can be written as $(2{t^2},4t)$ $...(ii)$
Now we know point $P(18,12)$ lies on the curve, therefore it must satisfy equation $(ii)$
i.e. $2{t^2} = 18$ and $4t = 12$
i.e. ${t^2} = \dfrac{{18}}{2}$ and $t = \dfrac{{12}}{4}$
i.e. ${t^2} = 9$ and $t = 3$
i.e. $t = \sqrt 9$ and $t = 3$
i.e. $t = \pm 3$ and $t = 3$
Therefore $t = 3$ $...(iii)$
Now for any other point $Q$ lying on the graph, the normal at $P$cuts it again at point $Q$say at point ${t_1}$ then, ${t_1} = (t + \dfrac{a}{t}) = (t + \dfrac{2}{t}) = (3 + \dfrac{2}{3}) = (\dfrac{{9 + 2}}{3}) = \dfrac{{11}}{3}$
Again by equation $(ii)$ coordinates of $Q$ are $\left( {2{{\left( {\dfrac{{11}}{3}} \right)}^2},4\left( {\dfrac{{11}}{3}} \right)} \right)$
$= \left( {2 \times \dfrac{{121}}{9},4 \times \dfrac{{11}}{3}} \right)$
$= \left( {\dfrac{{242}}{9},\dfrac{{44}}{3}} \right)$
Since, formula for length of a line joining two points $({x_1},{y_1})$ and $({x_2},{y_2})$ $= \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
Thus, length $PQ$ of the line joining the points $P(18,12)$ and Q\left( {\dfrac{{242}}{9},\dfrac{{44}}{3}} \right)$$$$ = \sqrt {{{\left( {18 - \dfrac{{242}}{9}} \right)}^2} - {{\left( {12 - \dfrac{{44}}{3}} \right)}^2}}
$= \sqrt {{{\left( {\dfrac{{18 \times 9 - 242}}{9}} \right)}^2} + {{\left( {\dfrac{{12 \times 3 - 44}}{3}} \right)}^2}}$
$= \sqrt {{{\left( {\dfrac{{ - 80}}{9}} \right)}^2} + {{\left( {\dfrac{{ - 8}}{3}} \right)}^2}}$
$= \sqrt {\left( {\dfrac{{6400}}{{81}}} \right) + \left( {\dfrac{{64}}{9}} \right)}$
$= \sqrt {\dfrac{{6400 + 64 \times 9}}{{81}}}$
$= \sqrt {\dfrac{{6400 + 576}}{{81}}}$
$= \sqrt {\dfrac{{6976}}{{81}}}$
$= \sqrt {\dfrac{{64 \times 109}}{{81}}}$
$= \sqrt {\dfrac{{{{(8)}^2} \times 109}}{{{{(9)}^2}}}}$
$= \dfrac{8}{9}\sqrt {109}$

Thus, $PQ = \dfrac{8}{9}\sqrt {109}$
i.e. $9PQ = 8\sqrt {109}$

Note:
In these types of problems, plotting the graph of parabola gives us an idea about the sign of the points. It is very common for students to get confused between a tangent and a normal. A tangent is a straight line that touches the parabola at one point but doesn’t cut the parabola, whereas a Normal is a straight line which is perpendicular to the tangent of the parabola.