Question

If the tangent to the parabola ${{y}^{2}}=4ax$ meets the axis in T and tangent at the vertex A in Y and the rectangle TAYG is completed, then the locus of G is
(a) ${{y}^{2}}+2ax=0$
(b) ${{y}^{2}}+ax=0$
(c) ${{x}^{2}}+ay=0$
(d) None of these

Answer 1

Hint:Suppose any point as $\left( a{{t}^{2}},2at \right)$ through which the tangent is drawn at ${{y}^{2}}=4ax$.Now, get the equation of tangent by T = 0, where we need to replace

\[{{y}^{2}}\to \dfrac{y+2at}{2}\]

$ x\to \dfrac{x+a{{t}^{2}}}{2} $

Now, get points Y and T by putting x = 0, y = 0 respectively to the equation by tangent calculated.Now, observe the coordinates of point F by using coordinates of Y and T.

Complete step-by-step answer:

Let $P\left( a{{t}^{2}},2at \right)$ be any point on the parabola ${{y}^{2}}=4ax$, in parametric form of coordinates.

As, we know tangent at any point $P\left( a{{t}^{2}},2at \right)$ to parabola ${{y}^{2}}=4ax$, ${{y}^{2}}-4ax=0$ can be given by

T = 0 …………….(i)

i.e. replace ${{y}^{2}}\to y\left( 2at \right)$

replace $x\to \left( \dfrac{x+a{{t}^{2}}}{2} \right)$

Now, the tangent at P meets the axis of the parabola in T and the tangent at vertex of parabola in y and hence, the rectangle TAYG is completed, and we need to determine the locus of point G with the help of the given information.

Now, we can observe that the y-axis is acting as a tangent at vertex (A).

A of the parabola, because y-axis is the line touching ${{y}^{2}}=4ax$ at one point only.

And as the axis of parabola ${{y}^{2}}=4ax$ is x-axis, so, T is the point at which tangent is intersecting the axis of parabola.

Hence, we can complete the rectangle.

TAYG to get point G.

Let us suppose the coordinates of point G is $\left( {{h}_{1}}k \right)$.

Tangent at point P can be given with the help of equation (i) as

Parabola: ${{y}^{2}}=4ax$

Tangent at P: $y\left( 2at \right)=\dfrac{4a\left( x+a{{t}^{2}} \right)}{2}$

$\Rightarrow 2ayt=2a\left( x+a{{t}^{2}} \right)$

$yt=x+a{{t}^{2}}$ ……………….(ii)

Now, we can get point “y’ by putting x = 0 to the equation (ii) and can get point y = 0 to the equation (ii), because the line of equation (ii) is intersecting at y on y-axis and at T on x-axis and we know x-coordinates on y-axis 0 and y-coordinates on x-axis is 0.

So, we get point ’y’ as

x – coordinates of y = 0

y-coordinates of ‘y:

put x = 0 to equation (ii)

\[yt=a{{t}^{2}}\]

y = at

s, point ‘y’: (0, at)

similarly,

y-coordinates of T = 0

x-coordinate of T :

put y = 0 in equation (ii)

\[0=x+a{{t}^{2}}\]

 \[x=-a{{t}^{2}}\]

Point T is given as $\left( -a{{t}^{2}},0 \right)$

And point A is given as (0, 0).

Now, as TAYG is a rectangle, it means AY = GT, GY = AT as the opposite sides of a rectangle are equal.

Now, the coordinates of point G will be the same as x-coordinate of T and y-coordinates of y because TAYG is a rectangle and AT = GY, AY = GY.

So, we get

$h=-a{{t}^{2}},$ k = at

Now, we can get value of t from the equation in k and t as

$t=\dfrac{k}{a}$

Now, put $t=\dfrac{k}{a}$in the equation with relations of h and t. so, we get

$h=-a\times {{\left( \dfrac{k}{a} \right)}^{2}}$

$ h=-a{{\dfrac{k}{{{a}^{2}}}}^{2}} $

$ h=-\dfrac{{{k}^{2}}}{a}, $

$ ah+{{k}^{2}}=0, $

 $ {{k}^{2}}+ah=0 $

Now, replace $\left( {{h}_{1}}k \right)\to $(x, y) to get the locus of point G as

${{y}^{2}}+ax=0$

So, option (b) is the correct answer.

Note: One may suppose point P as $\left( {{x}_{1}},{{y}_{1}} \right)$ as well i.e. not in parametric form of coordinates. We will get same result with this approach as well but we need to use one more equation i.e. ${{y}^{2}}_{1}=4a{{x}_{1}}$ to solve the problem. It means this approach would be longer. So, always try to use parametric coordinates with these kinds of questions.

One may suppose the direct tangent equation for parabola ${{y}^{2}}=4ax$ in slope form, which is given as $y=mx+\dfrac{a}{m}$ , now, apply the same approach to get T and Y and hence, get the locus of G.