Abscissae of points on the curve xy = (c + x) | vedclass

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

Question

(C) Given the curve $xy = (x + c)^2$,we can write $y = \frac{(x + c)^2}{x} = \frac{x^2 + 2cx + c^2}{x} = x + 2c + \frac{c^2}{x}$.Differentiating with

Vedclass · Accepted Answer

$\pm c / \sqrt{2}$

Vedclass · Answer

(C) Given the curve $xy = (x + c)^2$,we can write $y = \frac{(x + c)^2}{x} = \frac{x^2 + 2cx + c^2}{x} = x + 2c + \frac{c^2}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 - \frac{c^2}{x^2} = \frac{x^2 - c^2}{x^2}$.The slope of the normal at any point $(x, y)$ is $m_n = -\frac{dx}{dy} = -\frac{x^2}{x^2 - c^2} = \frac{x^2}{c^2 - x^2}$.The normal cuts off numerically equal intercepts from the axes,which means the slope of the normal must be $\pm 1$.Case $1$: $\frac{x^2}{c^2 - x^2} = 1 \Rightarrow x^2 = c^2 - x^2 \Rightarrow 2x^2 = c^2 \Rightarrow x = \pm \frac{c}{\sqrt{2}}$.Case $2$: $\frac{x^2}{c^2 - x^2} = -1 \Rightarrow x^2 = -(c^2 - x^2) \Rightarrow x^2 = -c^2 + x^2 \Rightarrow 0 = -c^2$,which is impossible for $c 
eq 0$.Thus,the abscissae are $\pm \frac{c}{\sqrt{2}}$.

Abscissae of points on the curve $xy = (c + x)^2$,the normal at which cuts off numerically equal intercepts from the axes of coordinates is/are:

Similar Questions

At which point on the $y$-axis is the line $x = 0$ a tangent to the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

Find the equation of the normal to the circle $x^2 + y^2 - 5x + 2y - 48 = 0$ at the point $(5, 6)$.

If $OA$ and $OB$ are the tangents from the origin to the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and $C$ is the centre of the circle,the area of the quadrilateral $OACB$ is

If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5}, 2)$ on the circle $x^{2}+y^{2}-2x-4y=0$ intersect at the point $Q$,then the area of the triangle $OPQ$ is equal to

If the line $3x - 4y = \lambda$ touches the circle $x^2 + y^2 - 4x - 8y - 5 = 0$,then $\lambda$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module