If the normal to the rectangular hyperbola xy | vedclass

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

Question

(B) The parametric coordinates of a point on the rectangular hyperbola $xy = c^2$ are $(ct, c/t)$.The slope of the tangent at point $t$ is given by $\

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) The parametric coordinates of a point on the rectangular hyperbola $xy = c^2$ are $(ct, c/t)$.The slope of the tangent at point $t$ is given by $\frac{dy}{dx} = -\frac{1}{t^2}$.Therefore,the slope of the normal at point $t$ is $m_N = t^2$.The equation of the normal at point $t$ is $y - \frac{c}{t} = t^2(x - ct)$,which simplifies to $y - \frac{c}{t} = t^2x - c t^3$,or $y = t^2x - c t^3 + \frac{c}{t}$.Since this normal meets the curve again at $t_1$,the point $(ct_1, c/t_1)$ must satisfy the equation of the normal:$\frac{c}{t_1} = t^2(ct_1) - ct^3 + \frac{c}{t}$Dividing by $c$ (assuming $c 
eq 0$):$\frac{1}{t_1} = t^2 t_1 - t^3 + \frac{1}{t}$$\frac{1}{t_1} - \frac{1}{t} = t^2 t_1 - t^3$$\frac{t - t_1}{t t_1} = -t^2(t - t_1)$Since $t 
eq t_1$ for the normal to meet the curve at a different point,we can divide by $(t - t_1)$:$\frac{1}{t t_1} = -t^2$$1 = -t^3 t_1$$t^3 t_1 = -1$.

If the normal to the rectangular hyperbola $xy = c^2$ at the point $t$ meets the curve again at $t_1$,then $t^3 t_1$ has the value equal to

Similar Questions

Let $x^2=4ky, k>0$ be a parabola with vertex $O(0,0)$. Let $BC$ be its latus rectum. An ellipse with center $P$ on $BC$ touches the parabola at $O$,and cuts $BC$ at points $D$ and $E$ such that $BD=DE=EC$ ($B, D, E, C$ in that order). The eccentricity of the ellipse is

If the curves $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ and $y^3 = 16x$ intersect at right angles,then $a^2 =$

The points of intersection of the curves whose parametric equations are $x = t^2 + 1, y = 2t$ and $x = 2s, y = \frac{2}{s}$ is given by

If $e_{1}$ and $e_{2}$ are the eccentricities of the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{4}=1$ and the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ respectively,and $(e_{1}, e_{2})$ is a point on the ellipse $15x^{2}+3y^{2}=k$,then $k$ is equal to:

Let a line $L: 2x + y = k, k > 0$ be a tangent to the hyperbola $x^2 - y^2 = 3$. If $L$ is also a tangent to the parabola $y^2 = \alpha x$,then $\alpha$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module