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.