Question

If two normal to a parabola \[{y^2} = 4ax\] intersect at right angles then the chord joining their feet passes through a fixed point whose coordinates are:
A.(-2a, 0)
B.(a, 0)
C.(2a, 0)
D.None of these.

Answer 1

Hint: To solve this problem we need to know the parameters of parabola. That is \[x = a{t^2}\] and \[y = 2at\] . We find the slope of the two normal lines and using some relation we will obtain the required answer. We know that the equation of parabola \[{y^2} = 4ax\] is \[y = 0\] and axis along the positive x-axis.

Complete step-by-step answer:
Given, two normal intersect at right angles and we obtain a chord PQ as shown in the figure.

Here we need to find the point \[R(\alpha ,0)\] .
Now, using the parametric of parabola we have points: \[P(at_1^2,2a{t_1})\] and \[Q(at_2^2,2a{t_2})\]
We know the slope of the normal is \[ = - t\] .
Now, the slope of normal at \[{N_1} = - {t_1}\] .
Slope of normal at \[{N_2} = - {t_2}\] .
We know that if \[{N_1} \bot {N_2}\] their product will be -1.
 \[ \Rightarrow ( - {t_1})( - {t_2}) = - 1\]
 \[ \Rightarrow {t_1}{t_2} = - 1\]
 \[ \Rightarrow {t_2} = - \dfrac{1}{{{t_1}}}\]
Now using this we get \[P(at_1^2,2a{t_1})\] and \[Q\left( {\dfrac{a}{{t_1^2}}, - \dfrac{{2a}}{{{t_1}}}} \right)\]
From the figure we have that the slope of PR is equal to the slope of PQ.
Also we know slope \[m = \dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}\]
Using the coordinates \[P(at_1^2,2a{t_1})\] \[R(\alpha ,0)\] and \[Q(at_2^2,2a{t_2})\] .
 \[{m_{PR}} = {m_{PQ}}\]
 \[ \Rightarrow \dfrac{{2a{t_1} - 0}}{{at_1^2 - \alpha }} = \dfrac{{2a{t_1} - 2a{t_2}}}{{at_1^2 - at_2^2}}\]
Taking a common on the right hand side,
 \[ \Rightarrow \dfrac{{2a{t_1} - 0}}{{at_1^2 - \alpha }} = \dfrac{{a(2{t_1} - 2{t_2})}}{{a(t_1^2 - t_2^2)}}\]
 \[ \Rightarrow \dfrac{{2a{t_1} - 0}}{{at_1^2 - \alpha }} = \dfrac{{(2{t_1} - 2{t_2})}}{{(t_1^2 - t_2^2)}}\]
We know \[{a^2} - {b^2} = (a - b)(a + b)\]
 \[ \Rightarrow \dfrac{{2a{t_1} - 0}}{{at_1^2 - \alpha }} = \dfrac{{2({t_1} - {t_2})}}{{({t_1} - {t_2})({t_1} + {t_2})}}\]
  Cancelling the terms,
 \[ \Rightarrow \dfrac{{2a{t_1} - 0}}{{at_1^2 - \alpha }} = \dfrac{2}{{({t_1} + {t_2})}}\]
Cross multiplication,
 \[ \Rightarrow (2a{t_1} - 0)\left( {{t_1} + {t_2}} \right) = 2(at_1^2 - \alpha )\]
 \[ \Rightarrow 2at_1^2 + 2a{t_1}{t_2} = 2at_1^2 - 2\alpha \]
Cancelling terms,
 \[ \Rightarrow - 2\alpha = 2a{t_1}{t_2}\]
 \[ \Rightarrow - \alpha = a{t_1}{t_2}\]
But we have \[{t_1}{t_2} = - 1\]
 \[ \Rightarrow - \alpha = - a\]
 \[ \Rightarrow \alpha = a\]
Thus we have \[R(a,0)\] .
That is, if two normal to a parabola \[{y^2} = 4ax\] intersect at right angles then the chord joining their feet passes through a fixed point whose coordinates are \[(a,0)\] .
So, the correct answer is “Option B”.

Note: Always try to write the given word problem into a diagram as we did above, it will give us some ideas about what needs to be found. Remember that in this case the slope of tangent \[\dfrac{1}{t}\] is and the slope of normal is \[ - t\] . Remember the formula for finding the slopes between two points. Careful about the calculation part.