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Question: A tangent to \({y^2} = 4ax\) meets the X- axis at T and tangent at vertex A in P and the rectangle T...

A tangent to y2=4ax{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) y2+4ax=0{y^2} + 4ax = 0
B) y2+2ax=0{y^2} + 2ax = 0
C) y2=2ax{y^2} = 2ax
D) y2+ax=0{y^2} + ax = 0

Explanation

Solution

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+amy = mx + \dfrac{a}{m} where mm is the slope and am\dfrac{a}{m} is the y-intercept.

Complete step by step solution:
We are given that a tangent to y2=4ax{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(h,k)\left( {h,k} \right).
We are given that a tangent to y2=4ax{y^2} = 4ax meets X- axis at T, so the co-ordinates of the Point T be(h,0)\left( {h,0} \right)
Equation of the Tangent of a line is given by y=mx+amy = mx + \dfrac{a}{m} where mm is the slope and am\dfrac{a}{m} is the y-intercept.
⇒\Rightarrow Equation of the Tangent of a line at Point T is 0=hm+am0 = hm + \dfrac{a}{m}
⇒\Rightarrow Equation of the Tangent of a line at Point T is am=−hm\dfrac{a}{m} = - hm
Thus, we get m2=−ah{m^2} = - \dfrac{a}{h} ……………………………………………………………………………………………(1)\left( 1 \right)
We are given that a tangent at vertex A in P, so the co-ordinates of the Point P be (0,k)\left( {0,k} \right)
Equation of the Tangent of a line is given by y=mx+amy = mx + \dfrac{a}{m} where mm is the slope and am\dfrac{a}{m} is the y-intercept.
⇒\Rightarrow Equation of the Tangent of a line at Point P is k=0+amk = 0 + \dfrac{a}{m}
⇒\Rightarrow Equation of the Tangent of a line at Point P is k=amk = \dfrac{a}{m}
Thus, we get m=akm = \dfrac{a}{k} ………………………………………………………………………………………………..(2)\left( 2 \right)
By squaring on both the sides of the equation (2)\left( 2 \right), we get
⇒m2=a2k2\Rightarrow {m^2} = \dfrac{{{a^2}}}{{{k^2}}} …………………………………………………………………………………………………………….(3)\left( 3 \right)
Now, by equating the equation (1)\left( 1 \right) and equation (3)\left( 3 \right), we get
⇒−ah=a2k2\Rightarrow - \dfrac{a}{h} = \dfrac{{{a^2}}}{{{k^2}}}
By cancelling the term, we get
⇒−1h=ak2\Rightarrow - \dfrac{1}{h} = \dfrac{a}{{{k^2}}}
⇒k2=−ah\Rightarrow {k^2} = - ah
Since (h,k)\left( {h,k} \right) are the co-ordinates of (x,y)\left( {x,y} \right), we get
⇒y2=−ax\Rightarrow {y^2} = - ax
By rewriting the equation, we get
⇒y2+ax=0\Rightarrow {y^2} + ax = 0

Therefore, the locus of Q is y2+ax=0{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.