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**Hint:**First of all, write the given equation of ellipse ${{x}^{2}}+4{{y}^{2}}=16$ in general form of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ by dividing 16 on both the sides we get $\dfrac{{{x}^{2}}}{16}+\dfrac{4{{y}^{2}}}{16}=1\Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1$. Now, the parametric coordinates of the ellipse at the point P $\left( a\cos \theta ,b\sec \theta \right)$ where “a is 4 and b is 2”. Then write the equation of normal passing through point P and to find the coordinates of point Q, put the y – coordinate of the equation of normal as 0 and find the value of x coordinate. Now, take the midpoint of P and Q to get the coordinates of M. Then find the locus of the midpoint which is the equation in x and y. After that, we are going to make the intersection of this locus with the latus rectum of the ellipse.

**Complete answer:**

The equation of ellipse given in the above problem is equal to:

${{x}^{2}}+4{{y}^{2}}=16$

We know that the general form of equation of ellipse is equal to:

$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Now, we are going to write the given equation of ellipse in the above general form we get,

$\begin{align}

& \dfrac{{{x}^{2}}}{16}+\dfrac{4{{y}^{2}}}{16}=1 \\

& \Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1 \\

\end{align}$

Now, we are going to write the parametric coordinates of P on the ellipse. We know that parametric coordinates on ellipse is equal to:

$\left( a\cos \theta ,b\sin \theta \right)$

In the above problem, “a is 4 and b is 2” so substituting these values of “a and b” in the above we get,

$\left( 4\cos \theta ,2\sin \theta \right)$

Now, we know that the parametric equation of normal on ellipse at point $\left( a\cos \theta ,b\sin \theta \right)$ is equal to:

$ax\sec \theta -by\text{cosec}\theta ={{a}^{2}}-{{b}^{2}}$

Now, substituting “a as 4 and b as 2” in the above equation we get,

$\begin{align}

& 4x\sec \theta -2y\text{cosec}\theta ={{\left( 4 \right)}^{2}}-{{\left( 2 \right)}^{2}} \\

& \Rightarrow 4x\sec \theta -2y\text{cosec}\theta =16-4 \\

& \Rightarrow 4x\sec \theta -2y\text{cosec}\theta =12 \\

\end{align}$

Dividing 2 on both the sides we get,

$\begin{align}

& \dfrac{4x\sec \theta -2y\text{cosec}\theta }{2}=\dfrac{12}{2} \\

& \Rightarrow 2x\sec \theta -y\text{cosec}\theta =6.......Eq.(1) \\

\end{align}$

Now, the normal from point P cuts the x axis at point Q which we are showing in the below figure:

In the above figure, we have shown the x and y axis by FG and EH respectively. Now, we are going to find the coordinates of point Q by putting y as 0 in eq. (1) we get,

$\begin{align}

& 2x\sec \theta -\left( 0 \right)\text{cosec}\theta =6 \\

& \Rightarrow 2x\sec \theta =6 \\

& \Rightarrow \dfrac{2x}{\cos \theta }=6 \\

& \Rightarrow x=3\cos \theta \\

\end{align}$

Hence, we got the coordinates of point Q as $(3\cos \theta ,0)$. Now, we are going to find the midpoint of PQ by adding x coordinates of P and Q and then divide them by 2. Similarly, we will find the coordinates of y. The midpoint of PQ is denoted by M.

$\begin{align}

& M\left( \dfrac{4\cos \theta +3\cos \theta }{2},\dfrac{2\sin \theta }{2} \right) \\

& =M\left( \dfrac{7\cos \theta }{2},\sin \theta \right) \\

\end{align}$

Now, writing the coordinates of Q and M in the above figure we get,

$\begin{align}

& 2x\sec \theta -\left( 0 \right)\text{cosec}\theta =6 \\

& \Rightarrow 2x\sec \theta =6 \\

& \Rightarrow \dfrac{2x}{\cos \theta }=6 \\

& \Rightarrow x=3\cos \theta \\

\end{align}$

Hence, we got the coordinates of point Q as $(3\cos \theta ,0)$. Now, we are going to find the midpoint of PQ by adding x coordinates of P and Q and then divide them by 2. Similarly, we will find the coordinates of y. The midpoint of PQ is denoted by M.

$\begin{align}

& M\left( \dfrac{4\cos \theta +3\cos \theta }{2},\dfrac{2\sin \theta }{2} \right) \\

& =M\left( \dfrac{7\cos \theta }{2},\sin \theta \right) \\

\end{align}$

Now, writing the coordinates of Q and M in the above figure we get,

Now, we are going to find the locus of the midpoint M by equating x coordinate as “h” and y coordinate as “k”.

$h=\dfrac{7}{2}\cos \theta ,k=\sin \theta $

In the above, rewriting the first equation we get,

$\begin{align}

& \dfrac{2h}{7}=\cos \theta ......Eq.(2) \\

& k=\sin \theta ........Eq.(3) \\

\end{align}$

We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so using this relation in the above equations by squaring both the sides of eq. (2) and eq. (3) and then adding them we get,

$\dfrac{4{{h}^{2}}}{49}+{{k}^{2}}=1$

Now, locus is a relation between x and y so substituting h as x and k as y in the above equation we get,

$\dfrac{4{{x}^{2}}}{49}+{{y}^{2}}=1$…… Eq. (4)

In the above problem, we have to find the intersection of this locus with the latus rectum. We know that the equation of latus rectum is equal to $x=\pm ae$. In this equation, we need to find “e” of the given ellipse:

We know that:

$e=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$

Substituting “b” as 2 and “a” as 4 in the above equation we get,

\[\begin{align}

& e=\sqrt{1-\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( 4 \right)}^{2}}}} \\

& \Rightarrow e=\sqrt{1-\dfrac{4}{16}} \\

& \Rightarrow e=\sqrt{\dfrac{16-4}{16}} \\

& \Rightarrow e=\dfrac{\sqrt{12}}{4} \\

& \Rightarrow e=\dfrac{2\sqrt{3}}{4} \\

& \Rightarrow e=\dfrac{\sqrt{3}}{2} \\

\end{align}\]

Now, substituting the value of “a” as 4 and “e” from the above in $x=\pm ae$ we get,

$\begin{align}

& x=\pm 4\left( \dfrac{\sqrt{3}}{2} \right) \\

& \Rightarrow x=\pm 2\sqrt{3} \\

\end{align}$

Substituting the above value of x in eq. (4) we get,

$\begin{align}

& \dfrac{4{{\left( 2\sqrt{3} \right)}^{2}}}{49}+{{y}^{2}}=1 \\

& \Rightarrow \dfrac{4\left( 12 \right)}{49}+{{y}^{2}}=1 \\

& \Rightarrow {{y}^{2}}=1-\dfrac{48}{49} \\

& \Rightarrow {{y}^{2}}=\dfrac{49-48}{49}=\dfrac{1}{49} \\

\end{align}$

Taking square root on both the sides we get,

$y=\pm \dfrac{1}{7}$

Hence, we got the intersection of the locus of the midpoint and latus rectum as $\left( \pm 2\sqrt{3},\pm \dfrac{1}{7} \right)$.

In the above problem, while taking the square root of ${{y}^{2}}=\dfrac{1}{49}$ on both the sides you might forget to write $\pm $ before $\dfrac{1}{7}$ but in this problem, all the options have $\pm $ sign so your answer won’t be wrong. But in this problem, you are having an advantage, it is not necessary that in every question you will get this advantage so make sure you won’t make this mistake.

$h=\dfrac{7}{2}\cos \theta ,k=\sin \theta $

In the above, rewriting the first equation we get,

$\begin{align}

& \dfrac{2h}{7}=\cos \theta ......Eq.(2) \\

& k=\sin \theta ........Eq.(3) \\

\end{align}$

We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so using this relation in the above equations by squaring both the sides of eq. (2) and eq. (3) and then adding them we get,

$\dfrac{4{{h}^{2}}}{49}+{{k}^{2}}=1$

Now, locus is a relation between x and y so substituting h as x and k as y in the above equation we get,

$\dfrac{4{{x}^{2}}}{49}+{{y}^{2}}=1$…… Eq. (4)

In the above problem, we have to find the intersection of this locus with the latus rectum. We know that the equation of latus rectum is equal to $x=\pm ae$. In this equation, we need to find “e” of the given ellipse:

We know that:

$e=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$

Substituting “b” as 2 and “a” as 4 in the above equation we get,

\[\begin{align}

& e=\sqrt{1-\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( 4 \right)}^{2}}}} \\

& \Rightarrow e=\sqrt{1-\dfrac{4}{16}} \\

& \Rightarrow e=\sqrt{\dfrac{16-4}{16}} \\

& \Rightarrow e=\dfrac{\sqrt{12}}{4} \\

& \Rightarrow e=\dfrac{2\sqrt{3}}{4} \\

& \Rightarrow e=\dfrac{\sqrt{3}}{2} \\

\end{align}\]

Now, substituting the value of “a” as 4 and “e” from the above in $x=\pm ae$ we get,

$\begin{align}

& x=\pm 4\left( \dfrac{\sqrt{3}}{2} \right) \\

& \Rightarrow x=\pm 2\sqrt{3} \\

\end{align}$

Substituting the above value of x in eq. (4) we get,

$\begin{align}

& \dfrac{4{{\left( 2\sqrt{3} \right)}^{2}}}{49}+{{y}^{2}}=1 \\

& \Rightarrow \dfrac{4\left( 12 \right)}{49}+{{y}^{2}}=1 \\

& \Rightarrow {{y}^{2}}=1-\dfrac{48}{49} \\

& \Rightarrow {{y}^{2}}=\dfrac{49-48}{49}=\dfrac{1}{49} \\

\end{align}$

Taking square root on both the sides we get,

$y=\pm \dfrac{1}{7}$

Hence, we got the intersection of the locus of the midpoint and latus rectum as $\left( \pm 2\sqrt{3},\pm \dfrac{1}{7} \right)$.

**Hence, the correct option is (c).****Note:**In the above problem, while taking the square root of ${{y}^{2}}=\dfrac{1}{49}$ on both the sides you might forget to write $\pm $ before $\dfrac{1}{7}$ but in this problem, all the options have $\pm $ sign so your answer won’t be wrong. But in this problem, you are having an advantage, it is not necessary that in every question you will get this advantage so make sure you won’t make this mistake.

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