The tangent to the hyperbola xy = c^2 at the | vedclass

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

Question

(D) The equation of the tangent to the hyperbola $xy = c^2$ at $P(ct, c/t)$ is $\frac{x}{ct} + \frac{yt}{c} = 2$.Setting $y=0$,we get $x=2ct$,so $T(2c

Vedclass · Accepted Answer

equal to $\frac{2}{c^2}$

Vedclass · Answer

(D) The equation of the tangent to the hyperbola $xy = c^2$ at $P(ct, c/t)$ is $\frac{x}{ct} + \frac{yt}{c} = 2$.Setting $y=0$,we get $x=2ct$,so $T(2ct, 0)$.Setting $x=0$,we get $y=\frac{2c}{t}$,so $T'(0, \frac{2c}{t})$.The equation of the normal at $P$ is $y - \frac{c}{t} = t^2(x - ct)$,which simplifies to $y = t^2x - ct^3 + \frac{c}{t}$.Setting $y=0$,we get $x = ct - \frac{c}{t^3}$,so $N(ct - \frac{c}{t^3}, 0)$.Setting $x=0$,we get $y = \frac{c}{t} - ct^3$,so $N'(0, \frac{c}{t} - ct^3)$.The area of $\Delta PNT$ is $\Delta = \frac{1}{2} |x_P(y_N - y_T) + x_N(y_T - y_P) + x_T(y_P - y_N)| = \frac{1}{2} |ct(0 - 0) + (ct - \frac{c}{t^3})(0 - \frac{c}{t}) + 2ct(\frac{c}{t} - 0)| = \frac{1}{2} |-\frac{c^2}{t^2} + \frac{c^2}{t^4} + 2c^2| = \frac{c^2(t^4+1)}{2t^4}$.The area of $\Delta PN'T'$ is $\Delta' = \frac{1}{2} |x_P(y_{N'} - y_{T'}) + x_{N'}(y_{T'} - y_P) + x_{T'}(y_P - y_{N'})| = \frac{1}{2} |ct(\frac{c}{t} - ct^3 - \frac{2c}{t}) + 0 + 0| = \frac{1}{2} |ct(-\frac{c}{t} - ct^3)| = \frac{c^2(1+t^4)}{2}$.Thus,$\frac{1}{\Delta} + \frac{1}{\Delta'} = \frac{2t^4}{c^2(1+t^4)} + \frac{2}{c^2(1+t^4)} = \frac{2(t^4+1)}{c^2(1+t^4)} = \frac{2}{c^2}$.

Similar Questions

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes is:

If the point $(1,1)$ and the origin lie in the same region with respect to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{1} = 1$ $(a > 0)$,then the range of $a$ is

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 5, 0)$,the transverse axis is of length $8$.

For the variable $t,$ the locus of the points of intersection of lines $x-2y=t$ and $x+2y=\frac{1}{t}$ is

The equation $\frac{x^2}{12 - \lambda} + \frac{y^2}{8 - \lambda} = 1$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module