If the tangent drawn at any point P a(cos4⁡θ,asin4⁡θ) on the curve x+y=a meets the co-ordinate axes in A, B respectively then.

The Curve: The given curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ is a standard form where $a$ is a constant. This represents a quarter-circle in the first quadrant of the Cartesian plane, with the radius $\sqrt{a}$ .
Point P on the Curve: The point $P(a \cos 4\theta, a \sin 4\theta)$ is assumed to lie on this curve. However, for it to lie on the curve, the relationship between $x$ and $y$ coordinates given by $\sqrt{x} + \sqrt{y} = \sqrt{a}$ should be satisfied. Let’s verify if $P$ does lie on the curve or not.
Tangent at Point P: If $P$ indeed lies on the curve, we will need the slope of the tangent line at $P$ to find where this line intersects the coordinate axes.

Let's start by verifying if

P(a \cos 4\theta, a \sin 4\theta)

lies on the curve

\sqrt{x} + \sqrt{y} = \sqrt{a}

.

Upon simplifying the expression, we find

\sqrt{a}(\sqrt{\sin(4\theta)^2} + \sqrt{\cos(4\theta)^2} - 1)

. Since

\sin^2\theta + \cos^2\theta = 1

for any

\theta

, the expression simplifies further to 0, indicating that point

P(a \cos 4\theta, a \sin 4\theta)

does indeed lie on the curve

\sqrt{x} + \sqrt{y} = \sqrt{a}

. Next, we need to find the slope of the tangent at

P

on the curve. This requires us to differentiate the curve

\sqrt{x} + \sqrt{y} = \sqrt{a}

implicitly to find

\frac{dy}{dx}

, and then use the chain rule to evaluate it at

P

. Let's proceed with this.

The expression for the slope of the tangent at point

P

simplifies to

-\frac{\sqrt{x}}{\sqrt{a} \sqrt{\sin(4\theta)^2}}

, but it should be evaluated further using the x-coordinate of

P

, which is

a \cos 4\theta

. Let's correct and evaluate this expression properly using the correct substitution.

The slope of the tangent line at point

P

on the curve simplifies to

-\frac{\sqrt{\cos(4\theta)^2}}{\sqrt{\sin(4\theta)^2}}

, which further simplifies to

-\frac{\cos 4\theta}{\sin 4\theta}

or

-\cot 4\theta

. This is the slope of the tangent line at

P

. With the slope

m = -\cot 4\theta

and the point

P(a \cos 4\theta, a \sin 4\theta)

, the equation of the tangent line can be written in point-slope form:

y - a \sin 4\theta = -\cot 4\theta (x - a \cos 4\theta)

If the tangent drawn at any point P $a (\cos^{4} θ, a \sin^{4} θ)$ on the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ meets the co-ordinate axes in A, B respectively then.