The abscissa of the point on the curve sqrt{x | vedclass

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

Question

(D) Given the curve equation: $\sqrt{xy} = a + x$. Squaring both sides,we get $xy = (a + x)^2 = a^2 + x^2 + 2ax$.Solving for $y$,we have $y = \frac{a^

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

Vedclass · Answer

(D) Given the curve equation: $\sqrt{xy} = a + x$. Squaring both sides,we get $xy = (a + x)^2 = a^2 + x^2 + 2ax$.Solving for $y$,we have $y = \frac{a^2}{x} + x + 2a$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{a^2}{x^2} + 1$.Since the tangent cuts off equal intercepts from the coordinate axes,its slope must be $\pm 1$. However,for a line to cut off equal intercepts $c$ and $c$ (where $c 
eq 0$),the slope is $-1$.Setting the derivative equal to $-1$: $-\frac{a^2}{x^2} + 1 = -1$.This simplifies to $\frac{a^2}{x^2} = 2$,which implies $x^2 = \frac{a^2}{2}$.Thus,$x = \pm \frac{a}{\sqrt{2}}$.Since $a > 0$,both values are valid abscissae.

The abscissa of the point on the curve $\sqrt{xy} = a + x$ where the tangent cuts off equal intercepts from the coordinate axes is $(a > 0)$.

Similar Questions

The curve $y=a x^3+b x^2+c x+5$ touches the $X$-axis at $P(-2,0)$. Then,$c=$

The point on the curve $y = 2 + \sqrt{4x + 1}$ at which the slope of the tangent is $\frac{2}{5}$ is:

For the curve $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2, (n \in N \text{ and } n > 1)$,the line $\frac{x}{a}+\frac{y}{b}=2$ is

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?

If the tangent at a point $P$ on the curve $y=4x^4+x$ is perpendicular to the tangent to the same curve at $(0,0)$,then the point $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module