Question

The normal to the parabola ${y^2} = 8x\;{\text{at}}\;(2,4)\;$ meets the parabola again at
A. $\left( {18,\;12} \right)$
B. $\left( {18,\; - 12} \right)$
C. $\left( { - 18,\;12} \right)$
D.None of these

Answer 1

Hint : To find the other end of the normal to the given parabola, first of all write the point in parametric form and then from the formula ${t_2} = - {t_1} - \dfrac{2}{{{t_1}}}$ find the value for the parameter for the end of the normal line by putting the value of ${t_1}$ . After getting the value parameter value for the end of the normal line, substitute it in the parametric form to get the required coordinate of the point.

Complete step-by-step answer :
In order to find the value of coordinates of the point where the normal to the parabola ${y^2} = 8x\;{\text{at}}\;(2,4)\;$ meets again, we have to first express the point in parametric form,
Parametric form of a point in a parabola ${y^2} = 4ax$ is given as $\left( {a{t^2},\;2at} \right)$
On comparing given equation with standard one, we get $a = 2$
Therefore the parametric point will be given as $\left( {2{t^2},\;4t} \right)$
So we have the point $(2,\;4) \equiv \left( {2{t^2},\;4t} \right) \Rightarrow 2 = 2{t^2}\;{\text{and}}\;4 = 4t$
On comparing we get value of $t = 1$
Now for point of normal at the other end of parabola is given as: ${t_2} = - {t_1} - \dfrac{2}{{{t_1}}},\;{\text{where}}\;{t_1}$ is the value of parametric parameter at first end, that is $t = 1$ in this question,
So,
${t_2} = - 1 - \dfrac{2}{1} = - 3$
Therefore required point will be given as $\left( {2{t_2}^2,\;4{t_2}} \right) \equiv \left( {2 \times {{( - 3)}^2},\;4 \times ( - 3)} \right) \equiv \left( {18,\; - 12} \right)$
So, the correct answer is “Option B”.

Note : We convert equations into parametric form in a way such that both the variables in the equation can be expressed with a single variable. Also we use parametric form to solve this type of question because parametric form has only one variable in comparison to standard form which has two.