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If a parabola whose length of latus rectum is 4a touches both coordinate axis then find the locus of its focus.
a) ${{x}^{2}}{{y}^{2}}={{a}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)$
b) ${{x}^{2}}{{y}^{2}}={{a}^{2}}\left( {{x}^{2}}-{{y}^{2}} \right)$
c) ${{x}^{2}}{{y}^{2}}={{a}^{3}}\left( {{x}^{3}}+{{y}^{2}} \right)$
d) None of these


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Last updated date: 30th May 2024
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Answer
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Hint: Consider the point on parabola in the form of $(a{{t}^{2}},2at)$ and draw tangents to the parabola. Then, we can use the distance between focus given by $(a,0)$ and tangents to formulate equations. The two tangents drawn to the parabola will be perpendicular to each other, so we can apply conditions for slopes of perpendicular lines to solve further.


Complete step-by-step solution -

It is given that the latus rectum is $4a$. So, let the given parabola ${{y}^{2}}=4ax$. Let the graph of parabola be like that which cuts both coordinates at A and B.

Let a point on parabola be $(a{{t}^{2}},2at)$. We have to find $\dfrac{dy}{dx}$ of the curve at a given point. Slope of tangent to given point $(a{{t}^{2}},2at)$ $=\dfrac{dy}{dx}$.
${{y}^{2}}=4ax$
We know $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$, so we get
$2y\dfrac{dy}{dx}=4a$
The given point is $(a{{t}^{2}},2at)$, so, $y=2at$
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{4a}{2(2at)} \\
 & \dfrac{dy}{dx}=\dfrac{1}{t} \\
\end{align}$
So the slope of tangent passes through point $(a{{t}^{2}},2at)$ is $=\dfrac{1}{t}\ldots \ldots \left( 1 \right)$
Now, we can substitute the points and the slope write the equation of tangent using the formula $y-{{y}_{1}}=m(x-{{x}_{1}})$ as
$\begin{align}
  & y-2at=\dfrac{1}{t}(x-a{{t}^{2}}) \\
 & t\left( y-2at \right)=x-a{{t}^{2}} \\
 & ty-2a{{t}^{2}}=x-a{{t}^{2}} \\
 & x-ty+a{{t}^{2}}=0 \\
\end{align}$
Let A $(a{{s}^{2}},2as)$ be a point on the parabola, on the x-axis. So, from the above discussion and using equation (1), we can conclude that the slope of tangent passing through $(a{{s}^{2}},2as)$ is $=\dfrac{1}{s}$. We can substitute it in the equation of tangent and we get $x-sy+a{{s}^{2}}=0$.
Now, we know the focus of parabola ${{y}^{2}}=4ax$ must be at $(a,0)$. Then the distance of focus from the tangent of the equation $x-sy+a{{s}^{2}}=0$ can be calculated using the formula of distance of a point from line. If we have a line $ax+by+c=0$, then the perpendicular distance from a point $(x,y)$ to line is given by $d=\left| \dfrac{xa+yb+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$.
Now let the distance from focus point $(a,0)$ be x. Then we have,
$\begin{align}
  & x=\left| \dfrac{a-0+a{{s}^{2}}}{\sqrt{{{1}^{2}}+{{s}^{2}}}} \right| \\
 & x=\left| \dfrac{a\left( 1+{{s}^{2}} \right)}{\sqrt{{{1}^{2}}+{{s}^{2}}}} \right| \\
 & x=\left| a\sqrt{{{1}^{2}}+{{s}^{2}}} \right|\ldots \ldots \ldots \left( 2 \right) \\
\end{align}$
Similarly, let B $(a{{r}^{2}},2ar)$ be a point on the parabola, on the y-axis. So, from the above discussion and using equation (1), we can conclude that the slope of tangent passing through $(a{{r}^{2}},2ar)$ is $=\dfrac{1}{r}$. We can substitute it in the equation of tangent and we get $x-ry+a{{r}^{2}}=0$.
Similarly, we can find the distance from focus point $(a,0)$ as $y$. So, we will get
$\begin{align}
  & y=\left| \dfrac{a-0+a{{r}^{2}}}{\sqrt{{{1}^{2}}+{{r}^{2}}}} \right| \\
 & y=\left| \dfrac{a\left( 1+{{r}^{2}} \right)}{\sqrt{{{1}^{2}}+{{r}^{2}}}} \right| \\
 & y=\left| a\sqrt{{{1}^{2}}+{{r}^{2}}} \right|\ldots \ldots \ldots \left( 3 \right) \\
\end{align}$
Now, we know that the tangents are perpendicular to each other. So, the product of their slope is equal to -1. Applying it, we will get
$\dfrac{1}{r}\times \dfrac{1}{s}=-1$
$rs=-1\ldots \ldots \ldots \left( 4 \right)$
Now using equation (2), we get
$\begin{align}
  & \dfrac{x}{a}=\left( \sqrt{{{1}^{2}}+{{s}^{2}}} \right) \\
 & {{\left( \dfrac{x}{a} \right)}^{2}}=\left( {{1}^{2}}+{{s}^{2}} \right) \\
 & {{\left( \dfrac{x}{a} \right)}^{2}}-1={{s}^{2}}\ldots \ldots \ldots \left( 5 \right) \\
\end{align}$
Now using equation (3), we get
$\begin{align}
  & \dfrac{y}{a}=\left( \sqrt{{{1}^{2}}+{{r}^{2}}} \right) \\
 & {{\left( \dfrac{y}{a} \right)}^{2}}=\left( {{1}^{2}}+{{r}^{2}} \right) \\
 & {{\left( \dfrac{y}{a} \right)}^{2}}-1={{r}^{2}}\ldots \ldots \ldots \left( 6 \right) \\
\end{align}$
Multiplying equation (5) and (6), we get
$\begin{align}
  & \left( {{\left( \dfrac{y}{a} \right)}^{2}}-1 \right)\left( {{\left( \dfrac{x}{a} \right)}^{2}}-1 \right)={{r}^{2}}{{s}^{2}} \\
 & \left( {{\dfrac{y}{{{a}^{2}}}}^{2}}-1 \right)\left( {{\dfrac{x}{{{a}^{2}}}}^{2}}-1 \right)={{r}^{2}}{{s}^{2}} \\
\end{align}$
Now, simplifying terms inside bracket and using equation (4), we get
$\begin{align}
  & \dfrac{{{x}^{2}}{{y}^{2}}}{{{a}^{2}}{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{a}^{2}}}+1={{(-1)}^{2}} \\
 & \dfrac{{{x}^{2}}{{y}^{2}}-{{a}^{2}}({{x}^{2}}+{{y}^{2}})}{{{a}^{4}}}+1=1 \\
 & \dfrac{{{x}^{2}}{{y}^{2}}-{{a}^{2}}({{x}^{2}}+{{y}^{2}})}{{{a}^{4}}}=0 \\
 & {{x}^{2}}{{y}^{2}}-{{a}^{2}}({{x}^{2}}+{{y}^{2}})=0 \\
 & {{x}^{2}}{{y}^{2}}={{a}^{2}}({{x}^{2}}+{{y}^{2}}) \\
\end{align}$
A parabola whose length of latus rectum is 4a touches both coordinate axis then the locus of its focus is ${{x}^{2}}{{y}^{2}}={{a}^{2}}({{x}^{2}}+{{y}^{2}})$.
So, the correct option must be a).

Note: Equation of a tangent to a parabola passing through point $(a{{t}^{2}},2at)$ can be taken as $x-ty+a{{t}^{2}}=0$ directly instead of deriving it. We must memorise this as it is very helpful in solving such questions and it will save time in the exam. In this question, latus rectum value is given through which we can find the focus of parabola and its equation.