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Question: From the point, where any normal to the parabola \( {y^2} = 4ac \) meets the axis, a line perpendicu...

From the point, where any normal to the parabola y2=4ac{y^2} = 4ac meets the axis, a line perpendicular to the normal is drawn. This line always touches the parabola
A. y2=4a(x−2a){y^2} = 4a\left( {x - 2a} \right)
B. y2+4a(x−2a)=0{y^2} + 4a\left( {x - 2a} \right) = 0
C. y2=4a(x+2a){y^2} = 4a\left( {x + 2a} \right)
D. y2+4a(x+2a)=0{y^2} + 4a\left( {x + 2a} \right) = 0

Explanation

Solution

Hint : The equation of any normal to the parabola is y=mx−2am−am3y = mx - 2am - a{m^3} . Then to find the straight line passing through the point perpendicular to the normal is y−y1=m(x−x1)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 y2=4ac{y^2} = 4ac meets at the axis.
The general equation of any normal to the parabola is,
⇒y=mx−am−am3\Rightarrow y = mx - am - a{m^3}
We know that the equation meets the axis in the point (2a+am2,0)\left( {2a + a{m^2},0} \right) .
Also, the equation to the straight line through the point (2a+am2,0)\left( {2a + a{m^2},0} \right) is perpendicular to the normal, that is given as,
⇒y=m1(x−2a−am2)\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 m1m=−1{m_1}m = - 1 .
Now, the above equation can be written as,
⇒y=m1(x−2a−am12)\Rightarrow y = {m_1}\left( {x - 2a - \dfrac{a}{{{m_1}^2}}} \right)
Hence, the equation we get is y=m1(x−2a)−am1y = {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 (2a,0)\left( {2a,0} \right) and whose concavity is always towards the non-positive end for the axis at the point xx . The equation is expressed as:
⇒y2=−4a(x−2a)\Rightarrow {y^2} = - 4a\left( {x - 2a} \right)
⇒y2+4a(x−2a)=0\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 yy axis is equal to slope times xx and subtract slope times aa and subtract aa 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 2a2a or 00 and the concavity is negative at the end of the xx axis.