Let a and b be positive real numbers such that a > 1 and b

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

normal

Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a$. Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$. Suppose the tangent to the hyperbola at $P$ passes through the point $(1,0)$, and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P$, the normal at $P$ and the $x$-axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are $TRUE$?

$(A)$ $1 < e < \sqrt{2}$

$(B)$ $\sqrt{2} < e < 2$

$(C)$ $\Delta=a^4$

$(D)$ $\Delta=b^4$

$A,D$

$A,B$

$A,C$

$B,D$

(IIT-2020)

Solution

Since Normal at point $P$ makes equal intercept on co-ordinate axes, therefore slope of Normal $=-1$

Hence slope of tangent $=1$

Equation of tangent

$y-0=1(x-1)$

$y=x-1$

Equation of tangent at $\left( x _1 y _1\right)$

$\frac{ xx _1}{ a ^2}-\frac{ yy _1}{ b ^2}=1$

$x-y=1$ (equation of Tangent)

on comparing $x _1= a ^2, y _1- b ^2$

Also $a^2-b^2=1$

Now equation of normal at $\left( x _1 y _1\right)=\left( a _1 b ^2\right)$

$y-b^2=-1\left(x-a^2\right)$

$x+y=a^2+b^2 \ldots \text { (Normal) }$

point of intersection with $x$-axis is $\left( a ^2+ b ^2\right)$

Now $e=\sqrt{1+\frac{b^2}{a^2}}$

$e=\sqrt{1+\frac{b^2}{b^2+1}} \quad\left[\text { from (1) } \frac{b^2}{b^2+1}<1\right] 1$

$\Delta=\frac{1}{2} \text {.AB.PQ }$

$\text { and } \Delta=\frac{1}{2}\left(a^2+b^2-1\right) \cdot b^2$

$\left.\Delta=\frac{1}{2}\left(2 b^2\right) b^2 \text { (from (1) } a^2-1=b^2\right)$

$\Delta=b^4 \text { so option (D) }$

Standard 11

Mathematics

Circles are drawn on chords of the rectangular hyperbola $ xy = c^2$ parallel to the line $ y = x $ as diameters. All such circles pass through two fixed points whose co-ordinates are :

normal

The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8,3 \sqrt{3})$ on it passes through the point

medium

(JEE MAIN-2022)

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

A point on the curve $\frac{{{x^2}}}{{{A^2}}} – \frac{{{y^2}}}{{{B^2}}} = 1$ is

normal

Consider the hyperbola

$\frac{x^2}{100}-\frac{y^2}{64}=1$

with foci at $S$ and $S_1$, where $S$ lies on the positive $x$-axis. Let $P$ be a point on the hyperbola, in the first quadrant. Let $\angle SPS _1=\alpha$, with $\alpha<\frac{\pi}{2}$. The straight line passing through the point $S$ and having the same slope as that of the tangent at $P$ to the hyperbola, intersects the straight line $S_1 P$ at $P_1$. Let $\delta$ be the distance of $P$ from the straight line $SP _1$, and $\beta= S _1 P$. Then the greatest integer less than or equal to $\frac{\beta \delta}{9} \sin \frac{\alpha}{2}$ is. . . . . . .

normal

(IIT-2022)

Solution

Similar Questions

Circles are drawn on chords of the rectangular hyperbola $ xy = c^2$ parallel to the line $ y = x $ as diameters. All such circles pass through two fixed points whose co-ordinates are :

The normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$ at the point $(8,3 \sqrt{3})$ on it passes through the point

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

A point on the curve $\frac{{{x^2}}}{{{A^2}}} – \frac{{{y^2}}}{{{B^2}}} = 1$ is

Start a Free Trial Now