Product of lengths of perpendiculars drawn from any point on hyperbola x^2 -2y^2 -2=0 to&nbsp

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10-2. Parabola, Ellipse, Hyperbola

normal

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

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{3}{2}$

$20$

Solution

Given $\frac{x^{2}}{2}-\frac{y^{2}}{1}=1$

Here, $a^{2}=2, \quad b^{2}=1$

Equation of asymptotes to the given hyperbola is

$\frac{x}{{\sqrt 2 }} – \frac{y}{1} = 0$ and ${\rm{ }}\frac{x}{{\sqrt 2 }} + \frac{y}{1} = 0$

Let $P(\sqrt{2} \sec \theta, \tan \theta)$ be any point, then product of length of perpendicular

${=\frac{\left[\frac{\sqrt{2} \sec \theta}{\sqrt{2}}-\frac{\tan \theta}{1}\right]\left[\frac{\sqrt{2} \sec \theta}{\sqrt{2}}+\frac{\tan \theta}{1}\right]}{\sqrt{\frac{1}{2}+\frac{1}{1}} \sqrt{\frac{1}{2}+\frac{1}{1}}}} $

${=\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\frac{3}{2}}=\frac{2}{3}}$

Standard 11

Mathematics

The eccentricity of the hyperbola conjugate to ${x^2} – 3{y^2} = 2x + 8$ is

medium

An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

$(A)$ Equation of ellipse is $x^2+2 y^2=2$

$(B)$ The foci of ellipse are $( \pm 1,0)$

$(C)$ Equation of ellipse is $x^2+2 y^2=4$

$(D)$ The foci of ellipse are $( \pm \sqrt{2}, 0)$

hard

(IIT-2009)

The asymptote of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}= 1$ form with any tangent to the hyperbola a triangle whose area is $a^2$ $\tan$ $ \lambda $ in magnitude then its eccentricity is :

normal

If the circle $x^2 + y^2 = a^2$ intersects the hyperbola $xy = c^2 $ in four points $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3), S(x_4, y_4), $ then

normal

Let $P$ is a point on hyperbola $x^2 -y^2 = 4$ , which is at minimum distance from $(0,-1)$ then distance of $P$ from $x-$ axis is

normal

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

Solution

Similar Questions

The eccentricity of the hyperbola conjugate to ${x^2} – 3{y^2} = 2x + 8$ is

The asymptote of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}= 1$ form with any tangent to the hyperbola a triangle whose area is $a^2$ $\tan$ $ \lambda $ in magnitude then its eccentricity is :

If the circle $x^2 + y^2 = a^2$ intersects the hyperbola $xy = c^2 $ in four points $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3), S(x_4, y_4), $ then

Let $P$ is a point on hyperbola $x^2 -y^2 = 4$ , which is at minimum distance from $(0,-1)$ then distance of $P$ from $x-$ axis is

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