Question

Prove that the sum of the intercepts of tangents to the curve $\sqrt x + \sqrt y = \sqrt a $ upon the axis of constant length.

Answer 1

Hint: A tangent to a circle is a straight line which touches the circles at only one point on its circumference. This point is called the point of tangency. The tangent to a circle is perpendicular to the radius at the point of tangency. The word tangent also has an important related meaning as a line or plane which touches a given curve at a single point.

Complete answer:
Given,
$\sqrt x + \sqrt y = \sqrt a $ …(1)
Differentiate the equation w.r.t. ’x’
\[ \Rightarrow \dfrac{1}{{2\sqrt x }} + \dfrac{1}{{2\sqrt y }}\dfrac{{dy}}{{dx}} = 0\]
Simplify the equation
\[ \Rightarrow \dfrac{1}{{2\sqrt x }} = - \dfrac{1}{{2\sqrt y }}\dfrac{{dy}}{{dx}}\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = - \sqrt {\dfrac{y}{x}} \]
Slope of the tangent is,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \sqrt {\dfrac{y}{x}} \]
Let $({x_1},{y_1})$be point on curve
Tangent at $({x_1},{y_1})$
As we know that formula of slope is
$ \Rightarrow (y - {y_0}) = m(x - {x_0})$
Put the value in the formula
$ \Rightarrow (y - {y_1}) = - \dfrac{{\sqrt y }}{{\sqrt x }}(x - {x_1})$
Simplify the equation
$ \Rightarrow \sqrt x (y - {y_1}) = - \sqrt y (x - {x_1})$
The intercept on y-axis
$x = 0$
$ \Rightarrow \sqrt x (y - {y_1}) = - \sqrt y (0 - {x_1})$
$ \Rightarrow y = \sqrt {{x_1}} \sqrt {{y_1}} + {y_1}$ …(2)
Intercepts on x-axis
$y = 0$
$ \Rightarrow \sqrt x (0 - {y_1}) = - \sqrt y (x - {x_1})$
$ \Rightarrow x = \sqrt {{x_1}} \sqrt {{y_1}} + {x_1}$ …(3)
Sum the equation 2 and 3 we get
$ \Rightarrow x + y = 2\sqrt {{x_1}} \sqrt {{y_1}} + {y_1} + {x_1}$
$ \Rightarrow x + y = {(\sqrt {{x_1}} + \sqrt {{y_1}} )^2}$
From equation 1 we know that
$\sqrt x + \sqrt y = \sqrt a $
So the following equation will become,
$ \Rightarrow x + y = {(\sqrt a )^2}$
$ \Rightarrow x + y = a$

Note:
Tangents are always perpendicular to the radius of the circle at the point of tangency. It never intersects the circle at the two points. The length of a tangent from an external point to a circle is equal. The angle between the chord and tangent is equal to the inscribed angle on the opposite side of that chord.