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# Find the points of intersection of the tangents at the end of the latus rectum of the parabola ${{y}^{2}}=4x$.

Last updated date: 11th Sep 2024
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Hint: We can compare a given parabola with the general equation of a parabola and obtain the length of the latus rectum as 4. Then we can use the distance formula to get the y coordinate of the ends of the latus rectum. Then we can use the equation of the tangent to the parabola at any point $\left( {{x}_{1}},{{y}_{1}} \right)$, which is given by $y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)$. Then, we can substitute the coordinates of the ends of the latus rectum to obtain two equations in x and y. Solving those, we will get the required answer.

Complete step-by-step solution -

The equation of the parabola is given in the question as ${{y}^{2}}=4x$.

Comparing it with the general equation, ${{y}^{2}}=4ax$ , we get that $a=1$.

For the parabola, ${{y}^{2}}=4ax$, the latus rectum is given by $x=a$ and the length of the latus rectum is $4a$. The focus is $\left( a,0 \right)$.

Therefore, for the parabola ${{y}^{2}}=4x$, we have the details as,

Latus rectum equation as $x=1$

Length of latus rectum $=4$

Focus at $\left( 1,0 \right)$

The figure below shows the details,

We have to compute the coordinates of the point $I$. For that, first, we have to find the coordinates of the ends of the latus rectum. The equation $x=1$ gives the $x$ coordinate of the latus rectum as $1$. Let the $y$ coordinate be $y$.

From the figure, it is clear that the distance $LS$ is $2$. So, using the distance formula,

$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$

$2=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 0-y \right)}^{2}}}$

$2=\sqrt{{{y}^{2}}}$

${{y}^{2}}=4$

$y=\pm 2$

Since the coordinates of the ends of the latus rectum, $L$ and $R$ are given by $\left( 1,y \right)$, we get it as $\left( 1,2 \right),\left( 1,-2 \right)$ respectively.

The equation of the tangent to parabola at any point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by

$y{{y}_{1}}=2a\left( x+{{x}_{1}} \right)\ldots \ldots \ldots \left( i \right)$

Therefore, the equation of the tangent to the end $L\left( 1,2 \right)$ of the latus rectum can be obtained by substituting $x=1,y=2,a=1$ in equation $\left( i \right)$ as,

$2y=2\left( x+1 \right)$

$y=\left( x+1 \right)$

$y-x=1\ldots \ldots \ldots \left( ii \right)$

The equation of the next tangent to the end $R\left( 1,-2 \right)$ of the latus rectum can be obtained by substituting $x=1,y=-2,a=1$ in equation $\left( i \right)$ as,

$-2y=2\left( x+1 \right)$

$-y=\left( x+1 \right)$

$-y-x=1\ldots \ldots \ldots \left( iii \right)$

Adding equations $\left( ii \right)$ and $\left( iii \right)$, we get

$y-x=1$

$-y-x=1$

$-2x=2$

$x=-1$

Substituting $x=-1$ in equation $\left( ii \right)$, we get

$y-(-1)=1$

$y+1=1$

$y=0$

Therefore, we get the point of intersection of the tangents at the ends of the latus rectum, $I$ of the parabola ${{y}^{2}}=4x$ as $\left( -1,0 \right)$.

The answer obtained is $\left( -1,0 \right)$.

Note: This question can be solved easily using the fact that the point of intersection of the tangents at the extremities of the latus rectum of a parabola is the foot of the directrix. For the parabola ${{y}^{2}}=4ax$, the directrix is given by $x=-a$. So, for the parabola ${{y}^{2}}=4x$ in this question, the directrix is $x=-1$. The coordinates of the foot of the directrix are hence $\left( -1,0 \right)$.