At which points on the curve sqrt{x} + sqrt{y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Differentiating the equation of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \

Vedclass · Accepted Answer

$(a, 0), (0, a), (a/4, a/4)$

Vedclass · Answer

(A) Differentiating the equation of the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ with respect to $x$,we get $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$,which simplifies to $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$.$(i)$ If the tangent is parallel to the $x$-axis,then $\frac{dy}{dx} = 0$. This implies $y = 0$. Substituting $y = 0$ into the curve equation,we get $\sqrt{x} = \sqrt{a}$,so $x = a$. The point is $(a, 0)$.$(ii)$ If the tangent is parallel to the $y$-axis,then $\frac{dy}{dx} = \infty$. This implies $x = 0$. Substituting $x = 0$ into the curve equation,we get $\sqrt{y} = \sqrt{a}$,so $y = a$. The point is $(0, a)$.$(iii)$ If the tangent makes equal angles with both axes,its slope $\frac{dy}{dx} = \pm 1$. Thus,$-\sqrt{\frac{y}{x}} = \pm 1$,which implies $\frac{y}{x} = 1$,or $y = x$. Substituting $y = x$ into the curve equation,we get $2\sqrt{x} = \sqrt{a}$,so $\sqrt{x} = \frac{\sqrt{a}}{2}$,which gives $x = \frac{a}{4}$. Since $y = x$,$y = \frac{a}{4}$. The point is $(\frac{a}{4}, \frac{a}{4})$.Thus,the points are $(a, 0), (0, a), (a/4, a/4)$.

At which points on the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$,is the tangent line: $(i)$ parallel to the $x$-axis? $(ii)$ parallel to the $y$-axis? $(iii)$ making equal angles with both axes?

Similar Questions

$x_1, x_2 \in N$. If a line having slope $2$ is a tangent to the curve $y=x^4-6x^3+13x^2-10x+5$ at points $P(x_1, y_1)$ and $Q(x_2, y_2)$, then $x_1x_2+y_1y_2=$

If the tangent drawn at the point $(\alpha, \beta)$ on the curve $x^{2/3} + y^{2/3} = 4$ is parallel to the line $\sqrt{3}x + y = 1$,then $\alpha^2 + \beta^2 =$

At any point for the curve $3y^2 = (x+5)^3$,if $ST$ represents the length of the subtangent and $SN$ represents the length of the subnormal,then $9(ST)^2 = $

Find the angle of intersection of the curves $y=4-x^{2}$ and $y=x^{2}$.

If the normal to the curve $y = f(x)$ at the point $(4, 6)$ makes an angle $\frac{2\pi}{3}$ with the positive $x$-axis in the anticlockwise direction,then $f'(4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module