Question

The x-intercept of the tangent at any arbitrary point of the curve $\dfrac{a}{{{x^2}}} + \dfrac{b}{{{y^2}}} = 1$ is proportional to
A. Square of the abscissa of the point of tangency
B. Square root of the abscissa of the point of tangency
C. Cube of the abscissa of the point of tangency
D. Cube root of the abscissa of the point of tangency

Answer 1

Hint: Given, curve $\dfrac{a}{{{x^2}}} + \dfrac{b}{{{y^2}}} = 1$, By doing the differentiation we’ll find the slope of the tangent at any point of the curve, using that we’ll find the equation of tangent using the one-point form of the line. Now substituting$y = 0$ we’ll get the x-intercept, which we’ll determine the proportionality, after finding its value.

Complete step by step answer:

Given data: $\dfrac{a}{{{x^2}}} + \dfrac{b}{{{y^2}}} = 1.........(i)$
$ \Rightarrow a{y^2} + b{x^2} = {x^2}{y^2}................(ii)$
On differentiating the given equation with-respect-to x, we get,
Using $\dfrac{{d{z^n}}}{{dx}} = n{z^{n - 1}}\dfrac{{dz}}{{dx}}$ , we get,
$ \Rightarrow \dfrac{{ - 2a}}{{{x^3}}} + \dfrac{{ - 2b}}{{{y^3}}}\dfrac{{dy}}{{dx}} = 0$
$ \Rightarrow \dfrac{{ - 2a}}{{{x^3}}} = \dfrac{{2b}}{{{y^3}}}\dfrac{{dy}}{{dx}}$
Multiplying both sides by $\dfrac{{{y^3}}}{{2b}}$, we get,
$ \Rightarrow \dfrac{{ - a{y^3}}}{{{x^3}b}} = \dfrac{{dy}}{{dx}}$
Now let an arbitrary point say \[\left( {l,m} \right)\]
Therefore the slope of the tangent at\[\left( {l,m} \right)\]i.e. $\dfrac{{dy}}{{dx}} = \dfrac{{ - a{m^3}}}{{{l^3}b}}$
We know that the equation of the line passing through \[\left( {{x_1},{y_1}} \right)\]and having slope ‘m’ is given by
$ \Rightarrow (y - {y_1}) = m(x - {x_1})$
Therefore the equation of the tangent at \[\left( {l,m} \right)\]is
 $ \Rightarrow (y - m) = \dfrac{{ - a{m^3}}}{{b{l^3}}}(x - l)$
For x-intercept \[y = 0\]
$ \Rightarrow - m = \dfrac{{ - a{m^3}}}{{b{l^3}}}(x - l)$
On Multiplying the equation by $\dfrac{{ - b{l^3}}}{{a{m^3}}}$, we get
$ \Rightarrow \dfrac{{b{l^3}}}{{a{m^2}}} = x - l$
$ \Rightarrow x = \dfrac{{b{l^3}}}{{a{m^2}}} + l$
Taking LCM and then taking ‘l’ common from both the terms
$ \Rightarrow x = l\left( {\dfrac{{b{l^2} + a{m^2}}}{{a{m^2}}}} \right)................(iii)$
Since \[\left( {l,m} \right)\]is point on the curve it will satisfy the curve
Therefore substituting the point\[\left( {l,m} \right)\]in equation(ii)
$ \Rightarrow a{m^2} + b{l^2} = {l^2}{m^2}$
On Substituting the value of $a{m^2} + b{l^2}$in equation(iii), we get,
$ \Rightarrow x = l\left( {\dfrac{{{l^2}{m^2}}}{{a{m^2}}}} \right)$
On cancelling the common terms we get,
$ \Rightarrow x = \dfrac{{{l^3}}}{a}$
Therefore we can say that x-intercept is proportional to the cube of abscissa
Hence, Option(C) is correct.

Note: X-intercept is the value of abscissa when any curve touches or passes through the x-axis i.e. we substitute $y = 0$ in the equation of tangent when we have to find the x-intercept similarly we can find the y-intercept of any curve by substituting $x = 0.$