Question

Find the length of the chord of the ellipse $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{{16}} = 1$, whose middle point is $\left( {\dfrac{1}{2},\dfrac{2}{5}} \right)$.

Answer 1

To solve the question first, we have to find out the equation of the chord. The intersection points of the chord and ellipse can be found out by solving the equations of the chord and ellipse. Finally, by using the distance formula we get the distance between two intersection points that is the length of the chord.

Complete step-by-step solution:
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We know that the general equation of an ellipse with semi major axis ’a’ and semi minor axis ‘b’ is given by
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ ………………….. (1)
From the question the equation of the ellipse is given by

\Rightarrow \dfrac{{{x^2}}}{{{5^2}}} + \dfrac{{{y^2}}}{{{4^2}}} = 1 \\\ $$ ……………………………… (2) We know the formula that the equation of the chord of an ellipse $$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$$ having midpoint is $$\left( {{x_{1,}}{y_1}} \right)$$given by $$\dfrac{{x{x_1}}}{{{a^2}}} + \dfrac{{y{y_1}}}{{{b^2}}} - 1 = \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} - 1$$ ………………………………….. (3) Applying this formula, the equation of the chord of $$\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{{16}} = 1$$having mid point $$\left( {\dfrac{1}{2},\dfrac{2}{5}} \right)$$ is given by $$ \dfrac{{x \times \dfrac{1}{2}}}{{{5^2}}} + \dfrac{{y \times \dfrac{2}{5}}}{{{4^2}}} - 1 = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{{{5^2}}} + \dfrac{{{{\left( {\dfrac{2}{5}} \right)}^2}}}{{{4^2}}} - 1 \\\ \Rightarrow \dfrac{x}{{50}} + \dfrac{y}{{40}} = \dfrac{1}{{100}} + \dfrac{1}{{100}} \\\ \Rightarrow 4x + 5y = 4 \\\ $$ ………………………………. (4) From eq. (4) we can deduce $$y = 4\left( {\dfrac{{1 - x}}{5}} \right)$$ ……………………………………. (5) $$ \Rightarrow \dfrac{{{x^2}}}{{25}} + \dfrac{{{{\left\\{ {4\left( {\dfrac{{1 - x}}{5}} \right)} \right\\}}^2}}}{{16}} = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{{25}} + \dfrac{{{{(1 - x)}^2}}}{{25}} = 1 \\\ \Rightarrow 2{x^2} - 2x - 24 = 0 \\\ \Rightarrow {x^2} - x - 12 = 0 \\\ \Rightarrow {x^2} - 4x + 3x - 24 = 0 \\\ \Rightarrow x(x - 4) + 3(x - 4) = 0 \\\ \Rightarrow (x - 4)(x + 3) = 0 \\\ \Rightarrow x = 4 or - 3 \\\ $$ …………………………………… (6) Substituting the value of x in Eq. (5) we get the values of y. For$$x = 4$$, $$y = 4\left( {\dfrac{{1 - 4}}{5}} \right) = - \dfrac{{12}}{5}$$ And for $$x = - 3$$, $$y = 4\left( {\dfrac{{1 + 3}}{5}} \right) = \dfrac{{16}}{5}$$ Therefore, the intersection points are $$\left( {4, - \dfrac{{12}}{5}} \right)$$ and $$\left( { - 3,\dfrac{{16}}{5}} \right)$$. Those are the coordinates of the end points of the chord. We know the distance between the points joining $$({x_1},{y_1})$$ and $$({x_2},{y_2})$$is given by $$d = \sqrt {{{({x_2} - {x_1})}^2} - {{({y_2} - {y_1})}^2}} $$ ……………………… (7) Hence applying this formula, we get the distance between the intersection points $$\left( {4, - \dfrac{{12}}{5}} \right)$$ and $$\left( { - 3,\dfrac{{16}}{5}} \right)$$ that is the length of the chord which is given by $$ d = \sqrt {{{( - 3 - 4)}^2} - {{\left( {\dfrac{{16}}{5} + \dfrac{{12}}{5}} \right)}^2}} \\\ = \dfrac{7}{5}\sqrt {41} \\\ $$ **Hence we got the length of the chord = $$\dfrac{7}{5}\sqrt {41} $$.** **Note:** While solving the equations of the chord and the ellipse we must get two coordinates since the degree of the equation of the ellipse is 2 and the obtained coordinates are the intersecting points. In an alternative method, to find the point of intersection of the chord and ellipse we can solve the equations for y and setting the expressions for y equal to each other we get values of x.