Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The sum of intercepts to the tangents to the curve $\sqrt x + \sqrt y = \sqrt a$ upon the coordinate axes is A. $2a$B. $a$C. $2\sqrt 2 a$D. None of these

Last updated date: 24th Jul 2024
Total views: 452.1k
Views today: 5.52k
Verified
452.1k+ views
Hint: Consider any point on the given curve and find the tangent to the curve at the considered point. Then take out the intercepts to the obtained tangent and add them, which gives the required solution.

Let $P\left( {{x_1},{y_1}} \right)$ be a point on the curve $\sqrt x + \sqrt y = \sqrt a$
Then, $\sqrt {{x_1}} + \sqrt {{y_1}} = \sqrt a .................................\left( i \right)$
Now, differentiating the curve $\sqrt x + \sqrt y = \sqrt a$, we have
$\Rightarrow \dfrac{{d\left( {\sqrt x } \right)}}{{dx}} + \dfrac{{d\left( {\sqrt y } \right)}}{{dx}} = \dfrac{{d\left( {\sqrt a } \right)}}{{dx}} \\ \Rightarrow \dfrac{1}{{2\sqrt x }} + \dfrac{1}{{2\sqrt y }}\dfrac{{dy}}{{dx}} = 0 \\ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sqrt y }}{{\sqrt x }} \\$
So, the slope of the tangent at $P\left( {{x_1},{y_1}} \right)$ to the curve $\sqrt x + \sqrt y = \sqrt a$ is
${\left( {\dfrac{{dy}}{{dx}}} \right)_P} = - \dfrac{{\sqrt {{y_1}} }}{{\sqrt {{x_1}} }}$
The equation of the tangent at $P\left( {{x_1},{y_1}} \right)$ to the curve $\sqrt x + \sqrt y = \sqrt a$ is
$y - {y_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_P}\left( {x - {x_1}} \right) \\ \Rightarrow y - {y_1} = - \dfrac{{\sqrt {{y_1}} }}{{\sqrt {{x_1}} }}\left( {x - {x_1}} \right) \\ \Rightarrow \left( {y - {y_1}} \right)\sqrt {{x_1}} = - \sqrt {{y_1}} \left( {x - {x_1}} \right) \\ \Rightarrow y\sqrt {{x_1}} - {y_1}\sqrt {{x_1}} = - x\sqrt {{y_1}} + {x_1}\sqrt {{y_1}} \\ \Rightarrow x\sqrt {{y_1}} + y\sqrt {{x_1}} = {x_1}\sqrt {{y_1}} + {y_1}\sqrt {{x_1}} \\$
Dividing both sides with $\sqrt {{x_1}} \sqrt {{y_1}}$ we have
$\Rightarrow \dfrac{x}{{\sqrt {{x_1}} }} + \dfrac{y}{{\sqrt {{y_1}} }} = \sqrt {{x_1}} + \sqrt {{y_1}}$
Since, $\sqrt {{x_1}} + \sqrt {{y_1}} = \sqrt a$
$\Rightarrow \dfrac{x}{{\sqrt {{x_1}} }} + \dfrac{y}{{\sqrt {{y_1}} }} = \sqrt a \\ \Rightarrow \dfrac{x}{{\sqrt a \sqrt {{x_1}} }} + \dfrac{y}{{\sqrt a \sqrt {{y_1}} }} = 1 \\ \Rightarrow \dfrac{x}{{\sqrt {a{x_1}} }} + \dfrac{y}{{\sqrt {a{y_1}} }} = 1 \\$
We know that if $\dfrac{x}{a} + \dfrac{y}{b} = 1$ is the equation of the line, then the intercepts upon the coordinate axes is $a{\text{ and }}b$.
So, the intercepts of the formed tangent $\dfrac{x}{{\sqrt {a{x_1}} }} + \dfrac{y}{{\sqrt {a{y_1}} }} = 1$ is $\sqrt {a{x_1}} {\text{ and }}\sqrt {a{y_1}}$.
Therefore, the sum of the intercepts is
$\Rightarrow \sqrt {a{x_1}} + \sqrt {a{y_1}} \\ \Rightarrow \sqrt a \left( {\sqrt {{x_1}} + \sqrt {{y_1}} } \right) \\$
Since, we have $\sqrt {{x_1}} + \sqrt {{y_1}} = \sqrt a$
$\Rightarrow \sqrt a \left( {\sqrt a } \right) \\ \Rightarrow a \\$
Therefore, the sum of the intercepts of the curve $\sqrt x + \sqrt y = \sqrt a$ upon the coordinate axes is $a$.
Thus, the correct option is B. $a$

Note: Here the tangent must touch both the coordinate axes to form intercepts. In the given above formula $a$ is the length of $x$ axis intercept and $b$ is the length of $y$ axis intercept.