Question

If the normal from any point to the parabola ${x^2} = 4y$ cuts the line y = 2 in points, whose abscissa are in A. P., then the slopes of the tangents at the three co – normal points are in:
A.A. P.
B.G. P.
C.H. P.
D.None of these

Answer 1

First, we will write the general co – ordinates on the parabola in parametric form as (2t, $t^2$) and then we will calculate the slope of tangent at any point T and then the slope of normal using the relation: slope of tangent$\times$slope of normal = -1. Then, we will write the equation of the normal at T as ($y$ – $y_1$) = m ($x$ – $x_1$) and upon simplification, we will get the equation of normal. We will assume any point P from where we have drawn the normal to the parabola and hence, the co – ordinates will satisfy the equation of normal (cubic equation). We will get three values as solution and hence three normal and all of them will pass through P. since the coefficient of $t^2$ is zero, hence sum of the roots will also be zero and then use the algebraic identity: if a + b+ c = 0, then
${a^3} + {b^3} + {c^3} = 3abc$, then we will put y = 2 as given In question, we will get t3 = x and then, we will calculate the points where y = 2 cuts the three normal and then if their abscissa is in A. P., then $2t_2^3 = t_1^3 + t_3^3$ and upon simplification, we will get the series in which the slopes of the tangents at three co – normal points are.

Complete step-by-step answer:
We are given a parabola ${x^2} = 4y$ and if a normal from any point to this parabola cuts the line y = 2 in points whose abscissa are in A.P., then we are required to find the nature of slopes of the tangent at the three co – normal points.
The general parametric form of co – ordinates of the parabola is (2t, t2).
Now, the slope of tangent at any point T will be $\dfrac{{dy}}{{dx}}$. On differentiating the equation of parabola with respect to x, we get
$\Rightarrow$ $\dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 4\dfrac{{dy}}{{dx}}$
$\Rightarrow 2x = 4\dfrac{{dy}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{2} \\\$
And on putting x = 2t, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = t$= mT
Now, we know that slope of tangent$\times$slope of normal = -1
$\Rightarrow$mT$\times$mN = -1
$\Rightarrow$mN = $\dfrac{{ - 1}}{t}$
Hence, we can write the equation of normal as: ($y$ – $y_1$) = m ($x$ –$x_1$) where can put $y_1$ =$t_2$ and $x_1$= $2t$
$\Rightarrow \left( {y{\text{ }}-{\text{ }}{{\text{t}}^2}} \right){\text{ }} = {\text{ }}\dfrac{{ - 1}}{t}{\text{ }}\left( {x{\text{ }}-{\text{ 2t}}} \right)$
$\Rightarrow ty - {t^3} = - x + 2t$
$\Rightarrow {t^3} - ty + 2t - x = 0 \\\ \Rightarrow {t^3} + t\left( {2 - y} \right) - x = 0 \\\$
Let the P (x1, y1) be the point from where the normal is drawn. Hence, the above equation will pass through this point and the point P will satisfy the above equation.
$\Rightarrow {t^3} + t\left( {2 - {y_1}} \right) - {x_1} = 0$
It is a cubic equation in t and let its roots be t1, t2 and t3.
So, from each of the roots, we will get a normal. The point corresponding to t1 will be A ($2t_1$, $t_1^2$), $t_2$ will be B ($2t_2$, $t_2^2$) and t3 will be C (2t3, $t_3^2$). If we draw normal to all three points, A, B and C, they will pass through P.
Here, we can see in equation (1) that the coefficient of t2 is zero and hence we can say that the sum of the roots is also zero.
$\Rightarrow {t_1} + {t_2} + {t_3} = 0$
Now, using the algebraic identity: if a + b+ c = 0, then ${a^3} + {b^3} + {c^3} = 3abc$
\Rightarrow $$$t_1^3 + t_2^3 + t_3^3 = 3{t_1}{t_2}{t_3}$$ We are given that the normal cut y = 2, putting this value in equation(1), we get
\Rightarrow {t^3} + t(2 - 2) - x = 0 \\
\Rightarrow {t^3} = x \\
$So, the points will become$\left( {t_1^3,2} \right),\left( {t_2^3,2} \right),\left( {t_3^3,2} \right)$and now, their abscissa is in A. P., then$ \Rightarrow 2t_2^3 = t_1^3 + t_3^3$Adding$t_2^3$both sides, we get$ \Rightarrow 3t_2^3 = t_1^3 + t_2^3 + t_3^3 Using the identity $$t_1^3 + t_2^3 + t_3^3 = 3{t_1}{t_2}{t_3}$$, we get \Rightarrow 3t_2^3 = 3{t_1}{t_2}{t_3}$$ \Rightarrow t_2^2 = {t_1}{t_3}$
Now, slope of tangent at A will be (mT)A = t1, at B: (mT)B and at C: (mT)C
Therefore, we can say that the given slopes are in G.P.
Hence, option(B) is correct.

Note: In this question, you may get confused at a lot of places as you can see that quite a number of formulae and concepts are used. You may go wrong while solving for the relation of the individual slopes and after that in the relation of their abscissas in A. P.
We have used algebraic identities which can be defined as: Identities (equality) which holds true for all values of the variables used in the respective identity. They are used in simplifying or re – arranging algebraic expressions (expressions which are formed using variables, integer constants and operations such as addition, subtraction, etc. )