Question

A tangent to ${y^2} = 4ax$ meets the X- axis at T and tangent at vertex A in P and the rectangle TAPQ is completed. Then the locus of Q is given by
A) ${y^2} + 4ax = 0$
B) ${y^2} + 2ax = 0$
C) ${y^2} = 2ax$
D) ${y^2} + ax = 0$

Answer 1

Hint:
Here, we will find the locus of the point Q which is the vertex of the rectangle. We will use the equation of the Tangent of the Point at T and P and by equating the equations of the Tangent, we will find the locus of the Point Q. Thus, the locus of the Point Q.

Formula Used:
Equation of the Tangent of a line is given by $y = mx + \dfrac{a}{m}$ where $m$ is the slope and $\dfrac{a}{m}$ is the y-intercept.

Complete step by step solution:
We are given that a tangent to ${y^2} = 4ax$ meets the X- axis at T and tangent at vertex A in P and the rectangle TAPQ is completed.
Let Q be a point$\left( {h,k} \right)$.
We are given that a tangent to ${y^2} = 4ax$ meets X- axis at T, so the co-ordinates of the Point T be$\left( {h,0} \right)$
Equation of the Tangent of a line is given by $y = mx + \dfrac{a}{m}$ where $m$ is the slope and $\dfrac{a}{m}$ is the y-intercept.
$ \Rightarrow $ Equation of the Tangent of a line at Point T is $0 = hm + \dfrac{a}{m}$
$ \Rightarrow $ Equation of the Tangent of a line at Point T is $\dfrac{a}{m} = - hm$
Thus, we get ${m^2} = - \dfrac{a}{h}$ ……………………………………………………………………………………………$\left( 1 \right)$
We are given that a tangent at vertex A in P, so the co-ordinates of the Point P be $\left( {0,k} \right)$
Equation of the Tangent of a line is given by $y = mx + \dfrac{a}{m}$ where $m$ is the slope and $\dfrac{a}{m}$ is the y-intercept.
$ \Rightarrow $ Equation of the Tangent of a line at Point P is $k = 0 + \dfrac{a}{m}$
$ \Rightarrow $ Equation of the Tangent of a line at Point P is $k = \dfrac{a}{m}$
Thus, we get $m = \dfrac{a}{k}$ ………………………………………………………………………………………………..$\left( 2 \right)$
By squaring on both the sides of the equation $\left( 2 \right)$, we get
$ \Rightarrow {m^2} = \dfrac{{{a^2}}}{{{k^2}}}$ …………………………………………………………………………………………………………….$\left( 3 \right)$
Now, by equating the equation $\left( 1 \right)$ and equation $\left( 3 \right)$, we get
$ \Rightarrow - \dfrac{a}{h} = \dfrac{{{a^2}}}{{{k^2}}}$
By cancelling the term, we get
$ \Rightarrow - \dfrac{1}{h} = \dfrac{a}{{{k^2}}}$
$ \Rightarrow {k^2} = - ah$
Since $\left( {h,k} \right)$ are the co-ordinates of $\left( {x,y} \right)$, we get
$ \Rightarrow {y^2} = - ax$
By rewriting the equation, we get
$ \Rightarrow {y^2} + ax = 0$

Therefore, the locus of Q is ${y^2} + ax = 0$ and thus Option (D) is the correct answer.

Note:
We know that slope is defined as the ratio of change in the y axis to the change in the x axis. Slope can be represented in the parametric form and in the point form. A point crossing the x-axis is called x-intercept and A point crossing the y-axis is called the y-intercept. We know that the tangent line can touch the circle exactly at one point. The tangent is a line which lies outside the circle or a curve.