Question

he equation to the locus of the middle point of the portion of the tangent to the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$ included between the coordinate axes is the curve-
A. $9{x^2} + 16{y^2} = 4{x^2}{y^2}$
B. $16{x^2} + 9{y^2} = 4{x^2}{y^2}$
C. $3{x^2} + 4{y^2} = 4{x^2}{y^2}$
D. $9{x^2} + 16{y^2} = {x^2}{y^2}$

Answer 1

Hint : The parametric form of tangent to the ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ is given as $\dfrac{{x\cos \theta }}{a} + \dfrac{{y\sin \theta }}{b} = 1$ at the point of contact $\left( {a\cos \theta ,b\sin \theta } \right)$
In this question the equation of the ellipse is given so by using the parametric form of tangent we will find the point of contact of tangent to ellipse through which the midpoint will be found through which we will form the equation of the locus.

Complete step-by-step answer :
Given the equation of the ellipse is $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$
Where major axis $a = 4$
Minor axis $b = 3$
Now let any tangent to the ellipse be $\dfrac{{x\cos \theta }}{4} + \dfrac{{y\sin \theta }}{3} = 1$
Now also assume that the tangent meets the ellipse at the point $A\left( {\dfrac{4}{{\cos \theta }},0} \right)$ and $B\left( {0,\dfrac{3}{{\sin \theta }}} \right)$
Let us assume that the point $\left( {h,k} \right)$ is the midpoint of A and B hence we can write
$2h = \dfrac{4}{{\cos \theta }}$ and $2k = \dfrac{3}{{\sin \theta }}$
These midpoints can also be written in the form
$\cos \theta = \dfrac{4}{{2h}}$
$\sin \theta = \dfrac{3}{{2k}}$
Now we know the general trigonometric equation ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , hence by substituting the values in the equation we can write

$${\left( {\dfrac{3}{{2k}}} \right)^2} + {\left( {\dfrac{4}{{2h}}} \right)^2} = 1 \\\ \dfrac{9}{{4{k^2}}} + \dfrac{{16}}{{4{h^2}}} = 1 \\\$$

By solving this equation we get
$16{k^2} + 9{h^2} = 4{h^2}{k^2}$
This equation can also be written as $16{y^2} + 9{x^2} = 4{x^2}{y^2}$
Hence we can say the equation of locus is $9{x^2} + 16{y^2} = 4{x^2}{y^2}$
So, the correct answer is “Option A”.

Note : Equation of tangent to an ellipse is written in slope form, point form and the parametric form. These forms are used to solve different types of problems for the tangent to an ellipse. Which form to be used has to be decided precisely.