Question

If A$\left( {4,0} \right)$ and B$\left( { - 4,0} \right)$ are two given points. A variable point P is such that PA+PB=10, then show that equation of locus of P is $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$.

Answer 1

Hint: In this type of question we will make use of distance between two points formula i.e., If A$({x_1},{y_1})$ and B$\left( {{x_2},{y_2}} \right)$ then the distance between these two points will be, AB=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ and also we also be using the square of difference of two variables i.e.,${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.

Complete answer:
Step 1:
Given two points are A$\left( {4,0} \right)$ and B$\left( { - 4,0} \right)$ and P is variable point such that PA+PB=10.
So, from the given data sum of distance between PA and PB is 10, and we have show that the locus of the point P will be $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$.
We have to use the point distance formula i.e., If A$({x_1},{y_1})$ and B$\left( {{x_2},{y_2}} \right)$ then the distance between these two points will be,
AB=$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.
Step 2:
Now here given that PA+PB=10, so let the point P be$\left( {x,y} \right)$.
PA+PB=10
By taking PB to R.H.S ,this can be rewritten as,
PA=10-PB----(1)
So, let's take L.H.S ,here PA$ = \sqrt {{{\left( {4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2}} $,
Now PB$ = \sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2}} $,
Step 3:
Now substituting these values in (1) we get,
$ \Rightarrow \sqrt {{{\left( {4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2}} = 10 - \sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( {0 - y} \right)}^2}} $
Now doing the subtraction inside the square root we get,
$ \Rightarrow \sqrt {{{\left( {4 - x} \right)}^2} + {{\left( y \right)}^2}} = 10 - \sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( y \right)}^2}} $
Now squaring on both sides we get,
$ \Rightarrow {\left( {\sqrt {{{\left( {4 - x} \right)}^2} + {{\left( y \right)}^2}} } \right)^2} = {\left( {10 - \sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( y \right)}^2}} } \right)^2}$
Now removing the square root on R.H.S and applying difference of two variables on L.H.S we get,
$ \Rightarrow {\left( {4 - x} \right)^2} + {y^2} = {10^2} + {\left( {\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} } \right)^2} - 2\left( {10} \right)\left( {\sqrt {{{\left( { - 4 - x} \right)}^2} + {{\left( y \right)}^2}} } \right)$
Now simplifying both the sides we get,
$ \Rightarrow 16 + {x^2} - 8x + {y^2} = 100 - 20\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} + {\left( { - 4 - x} \right)^2} + {y^2}$
Simplifying on L.H.S we get,
$ \Rightarrow 16 + {x^2} - 8x + {y^2} = 100 - 20\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} + 16 + 8x + {x^2} + {y^2}$
Now eliminating the like terms on both the sides we get,
$ \Rightarrow - 8x = 100 - 20\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} + 8x$,
Now taking all terms to one side we get,
$ \Rightarrow 0 = 100 - 20\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} + 8x + 8x$
Now again simplifying we get,
\[ \Rightarrow \dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1\]
Now take the square root term on L.H.S we get,
$ \Rightarrow 16x + 100 = 20\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} $
Now again squaring on both sides we get,
$ \Rightarrow {\left( {16x + 100} \right)^2} = {\left( {20\sqrt {{{\left( { - 4 - x} \right)}^2} + {y^2}} } \right)^2}$
Now applying the sum of squares of two variables on L.H.S and squaring on both sides we get,
\[ \Rightarrow 256{x^2} + 3200x + 10000 = \left( {400\left( {{{\left( { - 4 - x} \right)}^2} + {y^2}} \right)} \right)\]
Again simplifying the left hand side we get,
\[ \Rightarrow 256{x^2} + 3200x + 10000 = 400\left( {16 + {x^2} + 8x + {y^2}} \right)\]
Now multiplying 400 on R.H.S we get,
\[ \Rightarrow 256{x^2} + 3200x + 10000 = 6400 + 400{x^2} + 3200x + 400{y^2}\]
Now eliminating like terms on both the sides we get,
\[ \Rightarrow 256{x^2} + 10000 = 6400 + 400{x^2} + 400{y^2}\]
Now taking all \[x\]terms to R.H.S and constant term to L.H.S we get,
\[ \Rightarrow 10000 - 6400 = 400{x^2} + 400{y^2} - 256{x^2}\]
Now simplifying we get,
\[ \Rightarrow 144{x^2} + 400{y^2} = 3600\]
Now taking R.H.S to L.H.S we get,
\[ \Rightarrow \dfrac{{144{x^2}}}{{3600}} + \dfrac{{400{y^2}}}{{3600}} = 1\]
Now again simplifying we get,
\[ \Rightarrow \dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1\].
Thus the equation of locus of point P will be \[\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1\].
Hence Proved.

If A$\left( {4,0} \right)$ and B$\left( { - 4,0} \right)$ are two given points. A variable point P is such that PA+PB=10, then the equation of locus of P is$\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$.

Note:
The Distance Formula is a useful tool in finding the distance between two points which can be arbitrarily represented as points $({x_1},{y_1})$and$\left( {{x_2},{y_2}} \right)$, The Distance Formula itself is actually derived from the Pythagorean Theorem which is \[{a^2} + {b^2} = {c^2}\] where\[c\]is the longest side of a right triangle (also known as the hypotenuse) and \[a\]and\[b\] are the other shorter sides (known as the legs of a right triangle). The very essence of the Distance Formula is to calculate the length of the hypotenuse of the right triangle which is represented by the letter\[c\].