Question

The minimum radius vector of the curve $\dfrac{{{a}^{2}}}{{{x}^{2}}}+\dfrac{{{b}^{2}}}{{{y}^{2}}}=1$ is of length
A. $a-b$
B. $a+b$
C. $2a+b$
D. None of these.

Answer 1

In this problem, we have to find the length for the given minimum radius vector of the curve. We can first assume the radius vector as r. We can then find the minimum radius vector by differentiating the values of squared r. We can then simplify and solve for x and y and substitute in the suitable formula to get the required length.

Complete step by step solution:
We have to find the length for the given minimum radius vector of the curve $\dfrac{{{a}^{2}}}{{{x}^{2}}}+\dfrac{{{b}^{2}}}{{{y}^{2}}}=1$.
We can now assume the radius vector as ‘r’.
We know that ${{r}^{2}}={{x}^{2}}+{{y}^{2}}$…….. (1)
We can now write the given equation as,
$\Rightarrow {{x}^{2}}=\dfrac{{{a}^{2}}{{y}^{2}}}{{{y}^{2}}-{{b}^{2}}}$ …….. (2)
We can now substitute (2) in (1), we get
$\Rightarrow {{r}^{2}}=\dfrac{{{a}^{2}}{{y}^{2}}}{{{y}^{2}}-{{b}^{2}}}+{{y}^{2}}$ …… (3)
We know that for the minimum value of r we can find the value of $\dfrac{d\left( {{r}^{2}} \right)}{dy}=0$.
We can now differentiate (3), we get
$\Rightarrow \dfrac{d\left( {{r}^{2}} \right)}{dy}=\dfrac{\left( {{y}^{2}}-{{b}^{2}} \right){{a}^{2}}2y-{{a}^{2}}{{y}^{2}}\left( 2y \right)}{{{\left( {{y}^{2}}-{{b}^{2}} \right)}^{2}}}+2y$
We can now simplify the above step, we get
$\Rightarrow \dfrac{d\left( {{r}^{2}} \right)}{dy}=\dfrac{{{a}^{2}}2{{y}^{3}}-{{b}^{2}}{{a}^{2}}2y-{{a}^{2}}2{{y}^{3}}}{{{\left( {{y}^{2}}-{{b}^{2}} \right)}^{2}}}+2y$
We can now cancel the similar terms we get
$\Rightarrow \dfrac{d\left( {{r}^{2}} \right)}{dy}=\dfrac{-{{b}^{2}}{{a}^{2}}2y}{{{\left( {{y}^{2}}-{{b}^{2}} \right)}^{2}}}+2y=0$
We can now solve the above step, we get

& \Rightarrow \dfrac{-{{b}^{2}}{{a}^{2}}2y}{{{\left( {{y}^{2}}-{{b}^{2}} \right)}^{2}}}=-2y \\\ & \Rightarrow -2y{{b}^{2}}{{a}^{2}}=-2y{{\left( {{y}^{2}}-{{b}^{2}} \right)}^{2}} \\\ & \Rightarrow {{b}^{2}}{{a}^{2}}={{\left( {{y}^{2}}-{{b}^{2}} \right)}^{2}} \\\ \end{aligned}$$ We can now take square root on both sides, we get $$\begin{aligned} & \Rightarrow ab={{y}^{2}}-{{b}^{2}} \\\ & \Rightarrow {{y}^{2}}=b\left( a+b \right)......(4) \\\ \end{aligned}$$ Similarly, if we take y value instead of x in (2), we will get $$\Rightarrow {{x}^{2}}=a\left( a+b \right)$$ …….. (5) We can now substitute (4) and (5) in (1), we get $$\begin{aligned} & \Rightarrow {{r}^{2}}=a\left( a+b \right)+b\left( a+b \right) \\\ & \Rightarrow {{r}^{2}}=\left( a+b \right)\left( a+b \right)={{\left( a+b \right)}^{2}} \\\ & \Rightarrow r=a+b \\\ \end{aligned}$$ **Therefore, the required length is option B. $$a+b$$** **Note:** We should always remember that we can differentiate using the formula $$\dfrac{u}{v}=\dfrac{vu'-uv'}{{{v}^{2}}}$$. We should also remember the basic differentiation formulas. We should know that we can cancel a square root and the square term as the radical form represents half in the power.