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 distance between the directrices of a rectangular hyperbola is $10$ units. What is the distance between its foci?

The equation ${x^2} - 16xy - 11{y^2} - 12x + 6y + 21 = 0$ represents

The equation of the hyperbola whose foci are the same as the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and whose eccentricity is $2$ is:

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$.

Let $S$ be the focus of the hyperbola $x^2 - 2y^2 = 1$ lying on the positive $X$-axis. Let $P(-1, 1)$ be a given point. Then the area of the triangle formed by the line $PS$ with the coordinate axes is (in sq. units)

Vedclass Test Series

Exam Paper Generator

Online Exam Module