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

The graph of the conic $x^2 - (y - 1)^2 = 1$ has one tangent line with a positive slope that passes through the origin. If the point of tangency is $(a, b)$,then the length of the latus rectum of the conic is:

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

The equation of the tangents to the conic $3x^2 - y^2 = 3$ perpendicular to the line $x + 3y = 2$ is

In a hyperbola,if the length of the transverse axis is twice that of the conjugate axis,then the distance between its directrices is..... units.

The Cartesian form of the curve given by $x = \frac{a}{2} (t + \frac{1}{t})$ and $y = \frac{a}{2} (t - \frac{1}{t})$,where $t$ is a parameter,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module