Answer
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Hint: We start solving the problem by recalling the definition of the latus rectum of the parabola as the line segment joining two ends of parabola which passes through the focus and perpendicular to the axis of the parabola. We then draw the standard parabola and then find the equation of the parabola.
Using this equation, we find the ends of the latus rectum as they were the intersection points of latus rectum and parabola. We then find the length of the latus rectum which is equal to the distance between the two end points of the latus rectum.
Complete step by step answer:
According to the problem, we need to define the latus rectum of the parabola.
We know that the latus rectum is the line segment joining two ends of the parabola which passes through the focus and perpendicular to the axis of the parabola.
Let us draw the figure representing this definition and then find the equation and length of the latus rectum.
Let us assume that ${{y}^{2}}=4ax$ is the equation of the parabola. We know that the focus of the parabola ${{y}^{2}}=4ax$ is $S\left( a,0 \right)$, vertex of the parabola is $O\left( 0,0 \right)$ and axis of the parabola is $y=0$.
Now, let us assume the ends of the latus rectum are A and B.
Let us find the equation of the latus rectum of the parabola. We know that the equation of the line perpendicular to the $y=0$ is $x=b$ but we can see that the latus rectum passes through the point $S\left( a,0 \right)$. So, we get the equation of latus rectum as $x=a$.
Now, let us substitute $x=a$ in the equation of the parabola.
So, we get ${{y}^{2}}=4a\left( a \right)=4{{a}^{2}}$.
$\Rightarrow y=\pm 2a$.
So, the ends of the latus rectum will be $A\left( a,2a \right)$ and $B\left( a,-2a \right)$.
Now, let us find the length of the latus rectum which will be the distance between the points $A\left( a,2a \right)$ and $B\left( a,-2a \right)$.
So, the length of the latus rectum = $\sqrt{{{\left( a-a \right)}^{2}}+{{\left( -2a-2a \right)}^{2}}}$.
$\Rightarrow $ Length of the latus rectum = $\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -4a \right)}^{2}}}$.
$\Rightarrow $ Length of the latus rectum = $\sqrt{16{{a}^{2}}}$.
$\Rightarrow $ Length of the latus rectum = $4a$.
∴ The length of the latus rectum is $4a$.
Note: We need to know that latus rectum is a type of focal chord which is perpendicular to the axis of the parabola. We should know that the latus rectum is the only focal chord that has equal lengths above and below the axis of parabola. We should not confuse the vertex of the parabola with the focus of the parabola. Similarly, we can expect problems to find the angle subtended by the latus rectum at the vertex of parabola.
Using this equation, we find the ends of the latus rectum as they were the intersection points of latus rectum and parabola. We then find the length of the latus rectum which is equal to the distance between the two end points of the latus rectum.
Complete step by step answer:
According to the problem, we need to define the latus rectum of the parabola.
We know that the latus rectum is the line segment joining two ends of the parabola which passes through the focus and perpendicular to the axis of the parabola.
Let us draw the figure representing this definition and then find the equation and length of the latus rectum.
Let us assume that ${{y}^{2}}=4ax$ is the equation of the parabola. We know that the focus of the parabola ${{y}^{2}}=4ax$ is $S\left( a,0 \right)$, vertex of the parabola is $O\left( 0,0 \right)$ and axis of the parabola is $y=0$.
Now, let us assume the ends of the latus rectum are A and B.
Let us find the equation of the latus rectum of the parabola. We know that the equation of the line perpendicular to the $y=0$ is $x=b$ but we can see that the latus rectum passes through the point $S\left( a,0 \right)$. So, we get the equation of latus rectum as $x=a$.
Now, let us substitute $x=a$ in the equation of the parabola.
So, we get ${{y}^{2}}=4a\left( a \right)=4{{a}^{2}}$.
$\Rightarrow y=\pm 2a$.
So, the ends of the latus rectum will be $A\left( a,2a \right)$ and $B\left( a,-2a \right)$.
Now, let us find the length of the latus rectum which will be the distance between the points $A\left( a,2a \right)$ and $B\left( a,-2a \right)$.
So, the length of the latus rectum = $\sqrt{{{\left( a-a \right)}^{2}}+{{\left( -2a-2a \right)}^{2}}}$.
$\Rightarrow $ Length of the latus rectum = $\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -4a \right)}^{2}}}$.
$\Rightarrow $ Length of the latus rectum = $\sqrt{16{{a}^{2}}}$.
$\Rightarrow $ Length of the latus rectum = $4a$.
∴ The length of the latus rectum is $4a$.
Note: We need to know that latus rectum is a type of focal chord which is perpendicular to the axis of the parabola. We should know that the latus rectum is the only focal chord that has equal lengths above and below the axis of parabola. We should not confuse the vertex of the parabola with the focus of the parabola. Similarly, we can expect problems to find the angle subtended by the latus rectum at the vertex of parabola.
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