Question

Through each point of the straight line x = my + h is drawn the chord of the parabola ${{y}^{2}}=4ax$ which is bisected at the point. Prove that it always touches the parabola
${{\left( y-2am \right)}^{2}}=8a\left( x-h \right)$

Answer 1

Hint: Suppose a general coordinate \[\left( {{x}_{1}},{{y}_{1}} \right)\] on the line given in the problem. Draw a chord through, as the bisection of point of this chord is lying on the line. Use the property of polar with respect to mid-point of chord. Equation of chord will be same as the equation of polar with respect to the parabola. Equation of polar with respect to any point is given by equation
T = 0
Where, we need to replace
${{x}^{2}}\to x{{x}_{1}},{{y}^{2}}\to y{{y}_{1}},$
$x\to \dfrac{x+{{x}_{1}}}{2},y\to \dfrac{y+{{y}_{1}}}{2}$.

Complete step-by-step answer:
Let the equation of the parabola be ${{y}^{2}}=4ax$.
Now, it is given that the chords from the points on the straight-line x = my + h are drawn to the parabola, and hence, we need to prove that the lives will touch the parabola, if the lines are getting bisected at the point on the line x = my + h.
So, diagram with the help of above information can be given as

Let us suppose a general point on the line x = my + h.
As, we know, if $\left( {{x}_{1}},{{y}_{1}} \right)$ is the coordinates of the middle point of any chord, it will be parallel to the polar of $\left( {{x}_{1}},{{y}_{1}} \right)$with respect to the parabola. It is the property of polar for parabola.
Now, we know polar with respect to any point $\left( {{x}_{1}},{{y}_{1}} \right)$ can be given by the equation
T = 0……………(i)
Where, we need to use the equation of given curve and need to replace
$\begin{align}
  & {{x}^{2}}\to x{{x}_{1}} \\
 & {{y}^{2}}\to y{{y}_{1}} \\
 & x\to \dfrac{x+{{x}_{1}}}{2} \\
 & y\to \dfrac{y+{{y}_{1}}}{2} \\
\end{align}$
Hence, equation of polar w.r.t $\left( {{x}_{1}},{{y}_{1}} \right)$ for parabola ${{y}^{2}}=4ax$
$\begin{align}
  & y{{y}_{1}}=4a\left( \dfrac{x+{{x}_{1}}}{2} \right) \\
 & y{{y}_{1}}=2a\left( x+{{x}_{1}} \right) \\
\end{align}$
$y=\dfrac{2a}{{{y}_{1}}}\left( x+{{x}_{1}} \right)$……….(ii)
Now, we can compare the above equation with y = mx + c, so the slope of polar is $\dfrac{2a}{{{y}_{1}}}$.
As polar and the chord with mid-point $\left( {{x}_{1}},{{y}_{1}} \right)$are parallel. So, slopes of both the lines should be equal. Hence, we get slope of AB(chord) = $\dfrac{2a}{{{y}_{1}}}$
Now, as $\left( {{x}_{1}},{{y}_{1}} \right)$is lying on the line x = my + h as well. So, we can get the relation between ${{x}_{1}},{{y}_{1}}$ by satisfying the given point to the line. So, we get
${{x}_{1}}=m{{y}_{1}}+h$ ………………(iii)
It means the coordinates of M can be written as
$\left( m{{y}_{1}}+h,{{y}_{1}} \right)$
Hence, as equation of line with $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$as two points on it can be given as
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ …………………(iv)
Where m is the slope.
So, equation of chord can be given as
$\begin{align}
  & y-{{y}_{1}}=\dfrac{2a}{{{y}_{1}}}\left( x-\left( m{{y}_{1}}+h \right) \right) \\
 & y-{{y}_{1}}=\dfrac{2ax}{{{y}_{1}}}-\dfrac{2a}{{{y}_{1}}}\left( m{{y}_{1}}+h \right) \\
 & y-{{y}_{1}}=\dfrac{2ax}{{{y}_{1}}}-2am-\dfrac{2ah}{{{y}_{1}}} \\
 & y+2am=\dfrac{2a\left( x-h \right)}{{{y}_{1}}}+{{y}_{1}}, \\
\end{align}$
$\left( y+2am \right)=\dfrac{2a}{{{y}_{1}}}\left( x-h \right)+\dfrac{2a}{\left( \dfrac{2a}{{{y}_{1}}} \right)}$…………………..(v)
Now, as we know the tangent for the parabola ${{y}^{2}}=4ax$can be given by the equation
$y=mx+\dfrac{a}{m}$ …………….(vi)
Where the vertex of the parabola is (o, o).
But, with vertex at $\left( {{x}_{1}},{{y}_{1}} \right)$ for the parabola ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$, can be given as
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)+\dfrac{a}{m}$ ………………….(vii)
Now, we can compare equation (vii) and (v) and hence, get the vertex as
(h, -2am)
And value of a should be replaced by 2a in the equation of parabola as per the comparison.
So, equation of parabola is given as
${{\left( y+2am \right)}^{2}}=2\times 4a\left( x-h \right)$
${{\left( y+2am \right)}^{2}}=8a\left( x-h \right)$…………….(viii)
Hence, the chord will always touch the parabola represented in equation (viii).

Note: Property of polar with respect to the middle point of the chord makes the solution very flexible. One needs to clear this property to solve this question efficiently, otherwise, it will take a lot of time to get the slope and equation of the chord. Tangent $y=mx+\dfrac{a}{m}\to {{y}^{2}}=4ax$ can be proved by supposing a general equation of live in slope-intercept form i.e. y = mx + c. Satisfy
y = mx + c $\to {{y}^{2}}=4ax$as both will touch each other at one point only. It means the quadratic form by eliminating x, y will have only one solution i.e. discriminant $D={{b}^{2}}-4ac$ of that quadratic will be 0.