If x=9 is a chord of contact of the hyperbola | vedclass

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

Question

(B) Given the hyperbola $x^2-y^2=9$ and the chord of contact $x=9$. Substituting $x=9$ into the hyperbola equation: $81-y^2=9$ $\Rightarrow y^2=72$ $\

Vedclass · Accepted Answer

$3 x+2 \sqrt{2} y-3=0$

Vedclass · Answer

(B) Given the hyperbola $x^2-y^2=9$ and the chord of contact $x=9$. Substituting $x=9$ into the hyperbola equation: $81-y^2=9$ $\Rightarrow y^2=72$ $\Rightarrow y = \pm 6\sqrt{2}$. Thus,the points of contact are $P_1(9, 6\sqrt{2})$ and $P_2(9, -6\sqrt{2})$. Differentiating $x^2-y^2=9$ with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}$. At $P_1(9, 6\sqrt{2})$,the slope $m_1 = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$. The equation of the tangent at $P_1$ is $y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y - 24 = 3x - 27$,or $3x - 2\sqrt{2}y - 3 = 0$. At $P_2(9, -6\sqrt{2})$,the slope $m_2 = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$. The equation of the tangent at $P_2$ is $y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y + 24 = -3x + 27$,or $3x + 2\sqrt{2}y - 3 = 0$. Comparing with the options,$3x + 2\sqrt{2}y - 3 = 0$ is option $B$.

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

Similar Questions

$A$ hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3x^{2} + 4y^{2} = 12$. Then this hyperbola does not pass through which of the following points?

If $\frac{x^2}{\alpha+3}+\frac{y^2}{2-\alpha}=1$ represents a hyperbola,then $\alpha$ lies in

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

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$.

If $lx + my = 1$ is a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then $a^2 m^2 - b^2 l^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module