Question

From the point, where any normal to the parabola $ {y^2} = 4ac $ meets the axis, a line perpendicular to the normal is drawn. This line always touches the parabola
A. $ {y^2} = 4a\left( {x - 2a} \right) $
B. $ {y^2} + 4a\left( {x - 2a} \right) = 0 $
C. $ {y^2} = 4a\left( {x + 2a} \right) $
D. $ {y^2} + 4a\left( {x + 2a} \right) = 0 $

Answer 1

Hint: The equation of any normal to the parabola is $ y = mx - 2am - a{m^3} $ . Then to find the straight line passing through the point perpendicular to the normal is \[y - {y_1} = m\left( {x - {x_1}} \right)\].
where y1 is the y-axis point in the parabola, a is the focus point for parabola and m is the slope and x1 is the x-axis point in the parabola.

Complete step-by-step answer:
It is given that Parabola $ {y^2} = 4ac $ meets at the axis.
The general equation of any normal to the parabola is,
 $ \Rightarrow y = mx - am - a{m^3} $
We know that the equation meets the axis in the point $ \left( {2a + a{m^2},0} \right) $ .
Also, the equation to the straight line through the point $ \left( {2a + a{m^2},0} \right) $ is perpendicular to the normal, that is given as,
 $ \Rightarrow y = {m_1}\left( {x - 2a - a{m^2}} \right) $
It is known that the slope of the equation and normal to the equation is $ {m_1}m = - 1 $ .
Now, the above equation can be written as,
 $ \Rightarrow y = {m_1}\left( {x - 2a - \dfrac{a}{{{m_1}^2}}} \right) $
Hence, the equation we get is $ y = {m_1}\left( {x - 2a} \right) - \dfrac{a}{{{m_1}}} $ .
Then, we know that the straight line always touches the equal parabola, whose vertex is the point $ \left( {2a,0} \right) $ and whose concavity is always towards the non-positive end for the axis at the point $ x $ . The equation is expressed as:
 $ \Rightarrow {y^2} = - 4a\left( {x - 2a} \right) $
 $ \Rightarrow {y^2} + 4a\left( {x - 2a} \right) = 0 $
So, the correct answer is “Option B”.

Note: The other way to do the problem is the equation of any normal to the parabola is $ y $ axis is equal to slope times $ x $ and subtract slope times $ a $ and subtract $ a $ and slope cube. The straight line through the point perpendicular to the normal this straight line always touches the equal parabola whose vertex is always $ 2a $ or $ 0 $ and the concavity is negative at the end of the $ x $ axis.