Question

If the normal of the parabola ${{y}^{2}}=4x$ drawn at the end points of its latus rectum are tangents to the circle ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{r}^{2}}$, then what is the value of ${{r}^{2}}$.

Answer 1

To begin with, we will first define what a latus rectum of a parabola is and what are the end points at which it intersects the parabola. Then, we will find the ends of the latus rectum of the parabola ${{y}^{2}}=4x$ and find the slope of the tangents and thus normal at those points. Once we are able to find the slope, we can find the equation of normal at the two points. It is given that the normal are tangent to ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{r}^{2}}$. We know that the perpendicular distance from centre to any tangent is equal to radius of the circle. To understand the situation better, we shall also draw a diagram of various entities.

Complete step-by-step answer:
A latus rectum to a parabola is a line segment parallel to the directrix of the parabola at focus of the parabola.
Let ${{y}^{2}}=4ax$ be a parabola. Then the focus of the parabola will be at (a, 0). So, a latus rectum will be a line segment parallel to the directrix of the parabola and pass through (a, 0). We know that the directrix of the parabola ${{y}^{2}}=4ax$ is always perpendicular to x – axis. Thus, the latus rectum will also be perpendicular to x – axis and equation of latus rectum will be $x=a$.
Now, to find the points at which the latus rectum and parabola intersects, substitute $x=a$ in the equation ${{y}^{2}}=4ax$.
$\begin{aligned} & \Rightarrow {{y}^{2}}=4a\left( a \right) \\\ & \Rightarrow y=\pm 2a \\\ \end{aligned}$
Thus, the end two ends of the latus rectum are (a, 2a) and (a, ─2a).
Now, the equation of the parabola given to us is ${{y}^{2}}=4x$. As we can see, a = 1. Thus, the ends of the latus rectum will be (1, 2) and (1, ─2).
We shall now draw the diagram of all the entities in the question.

We will now find the slope of the tangent to the parabola at the ends of the latus rectum.
To find the slope at (1, 2) and (1, ─2), we need to differentiate the equation of parabola and substitute the points (1, 2) and (1, ─2).

& \Rightarrow \dfrac{d\left( {{y}^{2}}=4x \right)}{dx} \\\ & \Rightarrow 2y\dfrac{dy}{dx}=4 \\\ \end{aligned}$$ Now, substitute the y = 2 $\Rightarrow \dfrac{dy}{dx}=1$ and substitute y = ─2. $\Rightarrow \dfrac{dy}{dx}=-1$ Therefore, slope of tangents at points (1, 2) and (1, ─2) are 1 and ─1 respectively. We know that the product of the slope of perpendicular lines is ─1. Thus, slope of normal lines at (1, 2) and (1, ─2) are ─1 and 1 respectively. We can know form the normal line equations in slope point form $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, where $\left( {{x}_{1}},{{y}_{1}} \right)$ is a point on the line and m is the slope. Therefore, the equation of line passing through (1, 2) with slope ─1 is as follows. $\begin{aligned} & \Rightarrow y-2=-1\left( x-1 \right) \\\ & \Rightarrow y-2=-x+1 \\\ & \Rightarrow x+y=3 \\\ \end{aligned}$ An equation of line passing through (1, ─2) with slope 1 is as follows. $\begin{aligned} & \Rightarrow y+2=1\left( x-1 \right) \\\ & \Rightarrow y+2=x-1 \\\ & \Rightarrow x-y=3 \\\ \end{aligned}$ Now, to find the radius, we will find the perpendicular distance from the centre of the circle ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{r}^{2}}$ and line $x+y-3=0$. We know that the centre of the circle is (3, ─2). Perpendicular distance is found by the relation $d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$, where $ax+by+c=0$ is the line and $\left( {{x}_{1}},{{y}_{1}} \right)$ is the point. $\begin{aligned} & \Rightarrow d=\dfrac{\left| \left( 3 \right)+\left( -2 \right)-3 \right|}{\sqrt{1+1}} \\\ & \Rightarrow d=\dfrac{2}{\sqrt{2}} \\\ & \Rightarrow d=\sqrt{2} \\\ \end{aligned}$ Therefore, the radius of the circle is $\sqrt{2}$. Thus, the value of ${{r}^{2}}=2$. **Note:** Students can directly remember that the equation of the normal at any point $\left( a{{t}^{2}},2at \right)$ on the parabola ${{y}^{2}}=4ax$ is given by the equation $xt+y=2at+a{{t}^{3}}$ and the value of t for the ends of latus rectum are 1 and ─1.