Answer
Verified
419.4k+ views
Hint: To solve the question, we have to analyse that the point P on x-axis, implies a point P with y-coordinate equal to 0. Thus, the point P on the x-axis has only x-coordinate of P. To solve further, apply the formula for distance between two points to ease the procedure of solving and to arrive at the locus of point P.
Complete step-by-step answer:
Let the point P be (x, y).
We know the formula for the distance between two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given by \[\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\]
By applying the distance between two points formula for the points P(x, y) and (4, 0)
We get
\[\begin{align}
& \sqrt{{{\left( x-4 \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\
& =\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}} \\
\end{align}\]
Thus, the distance between two points formula for the points P(x, y) and (4, 0) \[=\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}}\]
The point on x-axis has no y-coordinate.
Thus, we get the point P(x, y) on the x-axis is (x, 0).
By applying the distance between two points formula for the points P(x, y) and (x, 0), we get
\[\begin{align}
& \sqrt{{{\left( x-x \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\
& =\sqrt{{{0}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{y}^{2}}} \\
& =y \\
\end{align}\]
Thus, the distance between two point’s formula for the points P(x, y) and (x, 0) \[=y\]
Given that the distance between P(x, y) and (4, 0) is twice the distance between the point P(x, y) and x-axis, we get
\[\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}}=2y\]
On squaring on both sides of the equation, we get
\[\begin{align}
& {{\left( \sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}} \right)}^{2}}={{\left( 2y \right)}^{2}} \\
& {{\left( x-4 \right)}^{2}}+{{y}^{2}}=4{{y}^{2}} \\
& {{\left( x-4 \right)}^{2}}=4{{y}^{2}}-{{y}^{2}} \\
& {{\left( x-4 \right)}^{2}}=3{{y}^{2}} \\
\end{align}\]
We know the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] . By applying the formula to the above equation, we get
\[\begin{align}
& {{x}^{2}}+{{4}^{2}}-2\times 4\times x=3{{y}^{2}} \\
& {{x}^{2}}+16-8x=3{{y}^{2}} \\
& {{x}^{2}}-3{{y}^{2}}-8x+16=0 \\
\end{align}\]
Thus, the locus of point P(x, y) is \[{{x}^{2}}-3{{y}^{2}}-8x+16=0\]
Note: The possibility of mistake can be, not applying the formula for distance between two points, which is important to ease the procedure of solving. The other possibility of mistake can be, not being able to analyse the point P on the x-axis, which implies a point P with y-coordinate equal to 0.
Complete step-by-step answer:
Let the point P be (x, y).
We know the formula for the distance between two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given by \[\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\]
By applying the distance between two points formula for the points P(x, y) and (4, 0)
We get
\[\begin{align}
& \sqrt{{{\left( x-4 \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\
& =\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}} \\
\end{align}\]
Thus, the distance between two points formula for the points P(x, y) and (4, 0) \[=\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}}\]
The point on x-axis has no y-coordinate.
Thus, we get the point P(x, y) on the x-axis is (x, 0).
By applying the distance between two points formula for the points P(x, y) and (x, 0), we get
\[\begin{align}
& \sqrt{{{\left( x-x \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\
& =\sqrt{{{0}^{2}}+{{y}^{2}}} \\
& =\sqrt{{{y}^{2}}} \\
& =y \\
\end{align}\]
Thus, the distance between two point’s formula for the points P(x, y) and (x, 0) \[=y\]
Given that the distance between P(x, y) and (4, 0) is twice the distance between the point P(x, y) and x-axis, we get
\[\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}}=2y\]
On squaring on both sides of the equation, we get
\[\begin{align}
& {{\left( \sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}} \right)}^{2}}={{\left( 2y \right)}^{2}} \\
& {{\left( x-4 \right)}^{2}}+{{y}^{2}}=4{{y}^{2}} \\
& {{\left( x-4 \right)}^{2}}=4{{y}^{2}}-{{y}^{2}} \\
& {{\left( x-4 \right)}^{2}}=3{{y}^{2}} \\
\end{align}\]
We know the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] . By applying the formula to the above equation, we get
\[\begin{align}
& {{x}^{2}}+{{4}^{2}}-2\times 4\times x=3{{y}^{2}} \\
& {{x}^{2}}+16-8x=3{{y}^{2}} \\
& {{x}^{2}}-3{{y}^{2}}-8x+16=0 \\
\end{align}\]
Thus, the locus of point P(x, y) is \[{{x}^{2}}-3{{y}^{2}}-8x+16=0\]
Note: The possibility of mistake can be, not applying the formula for distance between two points, which is important to ease the procedure of solving. The other possibility of mistake can be, not being able to analyse the point P on the x-axis, which implies a point P with y-coordinate equal to 0.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE