If x = 9 is the chord of contact of the hyper | vedclass

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

Question

(B) The equation of the chord of contact for a point $(h, k)$ with respect to the hyperbola $S: x^2 - y^2 - 9 = 0$ is given by $T = 0$,which is $xh -

Vedclass · Accepted Answer

$9x^2 - 8y^2 - 18x + 9 = 0$

Vedclass · Answer

(B) The equation of the chord of contact for a point $(h, k)$ with respect to the hyperbola $S: x^2 - y^2 - 9 = 0$ is given by $T = 0$,which is $xh - yk - 9 = 0$.Comparing this with the given chord $x = 9$ (or $x - 9 = 0$),we get $h = 1$ and $k = 0$.The equation of the pair of tangents from a point $(h, k)$ to the hyperbola is given by $SS_1 = T^2$.Here,$S = x^2 - y^2 - 9$,$S_1 = (1)^2 - (0)^2 - 9 = 1 - 9 = -8$,and $T = x(1) - y(0) - 9 = x - 9$.Substituting these into the formula: $(x^2 - y^2 - 9)(-8) = (x - 9)^2$.$-8x^2 + 8y^2 + 72 = x^2 - 18x + 81$.Rearranging the terms: $9x^2 - 8y^2 - 18x + 9 = 0$.

If $x = 9$ is the chord of contact of the hyperbola $x^2 - y^2 = 9$,then the equation of the corresponding pair of tangents is

Similar Questions

Let tangents drawn from point $C(0,-b)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ touch the hyperbola at points $A$ and $B$. If $\Delta ABC$ is a right-angled triangle,then $\frac{a^2}{b^2}$ is equal to -

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=c^2$ in four points $(x_i, y_i)$,for $i=1, 2, 3, 4$,then $y_1+y_2+y_3+y_4$ equals

The slope of the tangent drawn from the point $(1,1)$ to the hyperbola $2x^2-y^2=4$ is

If $y=mx+4$ ($n>0$ is a typo,likely $m>0$ or irrelevant) is a tangent to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$,then the point of contact of this tangent is

The directrix of the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module