Tangents are drawn to hyperbola frac{x^2}{9}-frac{y^2}{4}=1 , parallel to straight line 2 x-y=1 . po

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10-2. Parabola, Ellipse, Hyperbola

normal

Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, parallel to the straight line $2 x-y=1$. The points of contacts of the tangents on the hyperbola are

$(A)$ $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ $(B)$ $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

$(C)$ $(3 \sqrt{3},-2 \sqrt{2})$ $(D)$ $(-3 \sqrt{3}, 2 \sqrt{2})$

$(B,D)$

$(B,C)$

$(A,D)$

$(A,B)$

(IIT-2012)

Solution

Slope of tangents $=2$

Equation of tangents $y=2 x \pm \sqrt{9.4-4}$

$\Rightarrow y=2 x \pm \sqrt{32} $

$\Rightarrow 2 x-y \pm 4 \sqrt{2}=0$

Let point of contact be $\left( x _1, y _1\right)$ then equation $(i)$ will be identical to the equation

$\frac{x x_1}{9}-\frac{y_1}{4}-1=0 $

$\therefore \frac{x_1 / 9}{2}=\frac{y_1 / 4}{1}=\frac{-1}{ \pm 4 \sqrt{2}} $

$\Rightarrow\left(x_1, y_1\right) \equiv\left(-\frac{9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right) \text { and }\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

Standard 11

Mathematics

For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola $\mathrm{x}^2-\mathrm{y}^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is :

medium

(JEE MAIN-2024)

If the eccentricity of the standard hyperbola passing, through the point $(4, 6)$ is $2$, then the equation of the tangent to the hyperbola at $(4, 6)$ is

hard

(JEE MAIN-2019)

Equations of a common tangent to the two hyperbolas $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}$ $= 1 $ $\&$ $\frac{{{y^2}}}{{{a^2}}} – \frac{{{x^2}}}{{{b^2}}}$ $= 1 $ is :

normal

The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola)

normal

The point $\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $\frac{\sqrt{5}}{2} .$ If the tangent and normal at $\mathrm{P}$ to the hyperbola intersect its conjugate axis at the point $\mathrm{Q}$ and $\mathrm{R}$ respectively, then $QR$ is equal to :

hard

(JEE MAIN-2021)

Solution

Similar Questions

For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola $\mathrm{x}^2-\mathrm{y}^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is :

If the eccentricity of the standard hyperbola passing, through the point $(4, 6)$ is $2$, then the equation of the tangent to the hyperbola at $(4, 6)$ is

Equations of a common tangent to the two hyperbolas $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}$ $= 1 $ $\&$ $\frac{{{y^2}}}{{{a^2}}} – \frac{{{x^2}}}{{{b^2}}}$ $= 1 $ is :

The magnitude of the gradient of the tangent at an extremity of latera recta of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola)

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