The normal to the rectangular hyperbola xy = | vedclass

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

Question

(D) The equation of the normal to the rectangular hyperbola $xy = c^2$ at the point $t_1$ is given by $x t_1^2 - y = c(t_1^4 - 1)/t_1$.Since this norm

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) The equation of the normal to the rectangular hyperbola $xy = c^2$ at the point $t_1$ is given by $x t_1^2 - y = c(t_1^4 - 1)/t_1$.Since this normal meets the curve again at the point $t_2$,the point $(c t_2, c/t_2)$ must satisfy the equation of the normal.Substituting $x = c t_2$ and $y = c/t_2$ into the normal equation:$(c t_2) t_1^2 - c/t_2 = c(t_1^4 - 1)/t_1$Dividing by $c$:$t_2 t_1^2 - 1/t_2 = (t_1^4 - 1)/t_1$$(t_2^2 t_1^2 - 1)/t_2 = (t_1^4 - 1)/t_1$Since $t_1 
eq t_2$,we use the property that for a normal at $t_1$ meeting at $t_2$,the relation is $t_1^3 t_2 = -1$.

The normal to the rectangular hyperbola $xy = c^2$ at the point $t_1$ meets the curve again at the point $t_2$. Then the value of $t_1^3 t_2$ is

Similar Questions

The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4x+3y-7=0$ and $x-2y-1=0$ is

If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$,then the equation of the tangent at one of the points of contact is

The point of contact of the tangent $y = x + 2$ to the hyperbola $5x^2 - 9y^2 = 45$ is

If $A(4,0)$ and $B(-4,0)$ are two points,then the locus of a point $P$ such that $PA - PB = 4$ is

If the latus rectum subtends a right angle at the center of the hyperbola,then its eccentricity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module