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.