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(a). \[{{(2x-1)}^{2}}+4{{(y+1)}^{2}}=5\]

(b). \[{{(2x-1)}^{2}}+{{(y+1)}^{2}}=5\]

(c). \[{{(2x+1)}^{2}}+4{{(y-1)}^{2}}=5\]

(d). \[{{(2x-1)}^{2}}+4{{(y-1)}^{2}}=5\]

Answer
Verified

Hint:- In this question first we have to find the value of the point of intersection of the family of straight lines then apply Pythagoras theorem to get the required answer.

__Complete step-by-step solution -__

The equation of the family of lines is,

\[(4a+3)x-(a+1)y-(2a+1)=0\]

Expanding this, we get

\[4ax+3x-ay-y-2a-1=0\]

Now regrouping them, we get

\[3x-y-1+a(4x-y-2)=0\]

Now this is of the form \[{{L}_{1}}+\lambda {{L}_{2}}=0\].

Let the point of intersection of these two lines, i.e., \[3x-y-1=0\]and \[4x-y-2=0\]be P.

So, the given family of lines will pass through this fixed point P, as shown below.

Now we will find the coordinates of point P.

Consider,

\[3x-y-1=0\]

\[3x=y+1\]

\[x=\dfrac{y+1}{3}........(i)\]

Substitute this value in other equation of line, we get

\[4x-y-2=0\]

\[4\left( \dfrac{y+1}{3} \right)-y-2=0\]

\[\dfrac{4y+4-3y-(3\times 2)}{3}=0\]

\[y+4-6=0\]

\[y=6-4=2\]

Substituting the value of â€˜yâ€™ in equation (i), we get

\[x=\dfrac{2+1}{3}=\dfrac{3}{3}=1\]

So, the point of intersection of the family of lines is \[P(1,2)\].

Now, there is a perpendicular drawn from the origin onto each line. Let the foot of the perpendicular on each line be \[A(x,y)\], as shown below.

Now the points OPA will form a right-angled triangle in all the lines.

Applying Pythagoras theorem on \[\Delta OPA\], we get

\[O{{P}^{2}}=O{{A}^{2}}+A{{P}^{2}}.........(ii)\]

We know the distance between points \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\]is given by the distance formula, i.e.,

\[\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]

Applying the distance formula in equation (ii), we get

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

\[{{1}^{2}}+{{2}^{2}}={{x}^{2}}+{{y}^{2}}+{{(x-1)}^{2}}+{{(y-2)}^{2}}\]

Expanding, we get

\[1+4={{x}^{2}}+{{y}^{2}}+({{x}^{2}}+1-2x)+({{y}^{2}}+4-4y)\]

\[5=2{{x}^{2}}+2{{y}^{2}}+1-2x+4-4y\]

As this is not matching with the options, we will multiply the above equation with \[2\], we get

\[10=4{{x}^{2}}+4{{y}^{2}}+2-4x+8-8y\]

Grouping them, we get

\[4{{x}^{2}}-4x+2+4{{y}^{2}}-8y+8=10\]

Now we will try to convert this into \[{{(a+b)}^{2}}\]form, as in options.

\[\left( 4{{x}^{2}}-4x+1 \right)+1+4({{y}^{2}}-2y+1)+4=10\]

We know, \[\left( {{a}^{2}}-2ab+{{b}^{2}} \right)={{(a+b)}^{2}}\], so above equation can be re-written as,

\[{{\left( 2x-1 \right)}^{2}}+4{{(y-1)}^{2}}=10-4-1\]

\[{{\left( 2x-1 \right)}^{2}}+4{{(y-1)}^{2}}=5\]

So, the locus of the foot of the perpendicular from the origin on each member of the family \[(4a+3)x-(a+1)y-(2a+1)=0\] is \[{{\left( 2x-1 \right)}^{2}}+4{{(y-1)}^{2}}=5\].

Hence, the correct answer is option (d).

Note: Another method to find the answer is, instead of using Pythagoras theorem; we can find the slope of the equation of perpendicular line from the origin to a point. Then applying the formula, $y=mx+c$. But this would be a lengthy process.

The equation of the family of lines is,

\[(4a+3)x-(a+1)y-(2a+1)=0\]

Expanding this, we get

\[4ax+3x-ay-y-2a-1=0\]

Now regrouping them, we get

\[3x-y-1+a(4x-y-2)=0\]

Now this is of the form \[{{L}_{1}}+\lambda {{L}_{2}}=0\].

Let the point of intersection of these two lines, i.e., \[3x-y-1=0\]and \[4x-y-2=0\]be P.

So, the given family of lines will pass through this fixed point P, as shown below.

Now we will find the coordinates of point P.

Consider,

\[3x-y-1=0\]

\[3x=y+1\]

\[x=\dfrac{y+1}{3}........(i)\]

Substitute this value in other equation of line, we get

\[4x-y-2=0\]

\[4\left( \dfrac{y+1}{3} \right)-y-2=0\]

\[\dfrac{4y+4-3y-(3\times 2)}{3}=0\]

\[y+4-6=0\]

\[y=6-4=2\]

Substituting the value of â€˜yâ€™ in equation (i), we get

\[x=\dfrac{2+1}{3}=\dfrac{3}{3}=1\]

So, the point of intersection of the family of lines is \[P(1,2)\].

Now, there is a perpendicular drawn from the origin onto each line. Let the foot of the perpendicular on each line be \[A(x,y)\], as shown below.

Now the points OPA will form a right-angled triangle in all the lines.

Applying Pythagoras theorem on \[\Delta OPA\], we get

\[O{{P}^{2}}=O{{A}^{2}}+A{{P}^{2}}.........(ii)\]

We know the distance between points \[({{x}_{1}},{{y}_{1}})\]and \[({{x}_{2}},{{y}_{2}})\]is given by the distance formula, i.e.,

\[\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]

Applying the distance formula in equation (ii), we get

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

\[{{1}^{2}}+{{2}^{2}}={{x}^{2}}+{{y}^{2}}+{{(x-1)}^{2}}+{{(y-2)}^{2}}\]

Expanding, we get

\[1+4={{x}^{2}}+{{y}^{2}}+({{x}^{2}}+1-2x)+({{y}^{2}}+4-4y)\]

\[5=2{{x}^{2}}+2{{y}^{2}}+1-2x+4-4y\]

As this is not matching with the options, we will multiply the above equation with \[2\], we get

\[10=4{{x}^{2}}+4{{y}^{2}}+2-4x+8-8y\]

Grouping them, we get

\[4{{x}^{2}}-4x+2+4{{y}^{2}}-8y+8=10\]

Now we will try to convert this into \[{{(a+b)}^{2}}\]form, as in options.

\[\left( 4{{x}^{2}}-4x+1 \right)+1+4({{y}^{2}}-2y+1)+4=10\]

We know, \[\left( {{a}^{2}}-2ab+{{b}^{2}} \right)={{(a+b)}^{2}}\], so above equation can be re-written as,

\[{{\left( 2x-1 \right)}^{2}}+4{{(y-1)}^{2}}=10-4-1\]

\[{{\left( 2x-1 \right)}^{2}}+4{{(y-1)}^{2}}=5\]

So, the locus of the foot of the perpendicular from the origin on each member of the family \[(4a+3)x-(a+1)y-(2a+1)=0\] is \[{{\left( 2x-1 \right)}^{2}}+4{{(y-1)}^{2}}=5\].

Hence, the correct answer is option (d).

Note: Another method to find the answer is, instead of using Pythagoras theorem; we can find the slope of the equation of perpendicular line from the origin to a point. Then applying the formula, $y=mx+c$. But this would be a lengthy process.

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