Question

Find the locus of the middle points of chords of an ellipse whose distance from the centre is the constant length c.

Answer 1

In the above question, to find the locus of middle points of chord we will use a property of ellipse which states that T=S’, where T is the equation of chord when midpoint is given and S’ is the equation of ellipse which passes through that midpoint.

Complete step by step solution:
We know that the equation of ellipse is as under:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Let us assume the coordinates of midpoint (P) of the chord as h and k i.e. P(x,y)=P(h,k).
We know that when the midpoint is given and locus is asked we can use the relation T=S’ which basically is:
$\dfrac{xx'}{{{a}^{2}}}+\dfrac{yy'}{{{b}^{2}}}=\dfrac{{{\left( x' \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y' \right)}^{2}}}{{{b}^{2}}}$
In the above mentioned equation the x’ and y’ are the midpoints coordinates, so we will be substituting the midpoint coordinates in the above formula and we will get:
$\Rightarrow \dfrac{xh}{{{a}^{2}}}+\dfrac{yk}{{{b}^{2}}}=\dfrac{{{\left( h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( k \right)}^{2}}}{{{b}^{2}}}$
Now we will be taking LCM and making the denominator common and with this we get:
$\Rightarrow \dfrac{xh{{b}^{2}}+yk{{a}^{2}}}{{{a}^{2}}{{b}^{2}}}=\dfrac{{{\left( h \right)}^{2}}{{b}^{2}}+{{\left( k \right)}^{2}}{{a}^{2}}}{{{a}^{2}}{{b}^{2}}}$
As the denominators are equal on both RHS and LHS we can cancel them out and then we will get:
$\Rightarrow xh{{b}^{2}}+yk{{a}^{2}}={{\left( h \right)}^{2}}{{b}^{2}}+{{\left( k \right)}^{2}}{{a}^{2}}$
In the question it is also mentioned that the distance from center to the midpoint is c, by using the distance between two points which is:
$\left| \dfrac{xh{{b}^{2}}+yk{{a}^{2}}-{{\left( h \right)}^{2}}{{b}^{2}}-{{\left( k \right)}^{2}}{{a}^{2}}}{\sqrt{{{\left( {{b}^{2}}h \right)}^{2}}+{{\left( {{a}^{2}}k \right)}^{2}}}} \right|=c$
Now let’s assume the center to be lying at (0,0) so we can substitute (0,0) in the above distance equation and then calculate for the locus of the point.
After substituting (0,0) we got:

& \Rightarrow \left| \dfrac{\left( 0 \right)h{{b}^{2}}+\left( 0 \right)k{{a}^{2}}-{{\left( h \right)}^{2}}{{b}^{2}}-{{\left( k \right)}^{2}}{{a}^{2}}}{\sqrt{{{\left( {{b}^{2}}h \right)}^{2}}+{{\left( {{a}^{2}}k \right)}^{2}}}} \right|=c \\\ & \Rightarrow \left| \dfrac{-{{\left( h \right)}^{2}}{{b}^{2}}-{{\left( k \right)}^{2}}{{a}^{2}}}{\sqrt{{{\left( {{b}^{2}}h \right)}^{2}}+{{\left( {{a}^{2}}k \right)}^{2}}}} \right|=c \\\ \end{aligned}$$ Now we know that the value in modulus will always come out as positive so the negative sign in numerator will change into positive and we will get: $$\Rightarrow \dfrac{{{\left( h \right)}^{2}}{{b}^{2}}+{{\left( k \right)}^{2}}{{a}^{2}}}{\sqrt{{{\left( {{b}^{2}}h \right)}^{2}}+{{\left( {{a}^{2}}k \right)}^{2}}}}=c$$ Now we will be squaring both the sides i.e. RHS and LHS and we will get: $$\begin{aligned} & \Rightarrow {{\left( \dfrac{{{\left( h \right)}^{2}}{{b}^{2}}+{{\left( k \right)}^{2}}{{a}^{2}}}{\sqrt{{{\left( {{b}^{2}}h \right)}^{2}}+{{\left( {{a}^{2}}k \right)}^{2}}}} \right)}^{2}}={{c}^{2}} \\\ & \Rightarrow {{\left( {{\left( h \right)}^{2}}{{b}^{2}}+{{\left( k \right)}^{2}}{{a}^{2}} \right)}^{2}}={{c}^{2}}\left( {{\left( {{b}^{2}}h \right)}^{2}}+{{\left( {{a}^{2}}k \right)}^{2}} \right) \\\ & \Rightarrow {{\left( {{\left( h \right)}^{2}}{{b}^{2}}+{{\left( k \right)}^{2}}{{a}^{2}} \right)}^{2}}={{c}^{2}}\left( {{b}^{4}}{{h}^{2}}+{{a}^{4}}{{k}^{2}} \right) \\\ \end{aligned}$$ Now we will substitute the x and y in place of h and which were assumed coordinates of midpoint, and after substituting we will get the locus of the midpoint as: $${{\left( {{x}^{2}}{{b}^{2}}+{{y}^{2}}{{a}^{2}} \right)}^{2}}={{c}^{2}}\left( {{b}^{4}}{{x}^{2}}+{{a}^{4}}{{y}^{2}} \right)$$ So the required locus of the midpoint is $${{\left( {{x}^{2}}{{b}^{2}}+{{y}^{2}}{{a}^{2}} \right)}^{2}}={{c}^{2}}\left( {{b}^{4}}{{x}^{2}}+{{a}^{4}}{{y}^{2}} \right)$$ **Note:** The basic mistakes that happen in the above question is the application of the constant c is lost which will result in making the question complex, always make use of the things that are mentioned in the question like in this one, we can clearly see that the distance from the centre is mentioned as c and for that we used the formula distance between two points. Always take the centre as zero until it's given these kinds of problems to easily solve the questions.