The locus of a point P(alpha, beta) moving un | vedclass

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

Question

(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given the

Vedclass · Accepted Answer

$A$ hyperbola

Vedclass · Answer

(B) The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Given the line $y = \alpha x + \beta$,we have $m = \alpha$ and $c = \beta$.Substituting these into the condition,we get $\beta^2 = a^2\alpha^2 - b^2$.Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $y^2 = a^2x^2 - b^2$,which can be rewritten as $a^2x^2 - y^2 = b^2$.This equation represents a hyperbola.

The locus of a point $P(\alpha, \beta)$ moving under the condition that the line $y = \alpha x + \beta$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

Similar Questions

The length of the transverse axis of a hyperbola is $7$ and it passes through the point $(5, -2)$. The equation of the hyperbola is

The equation of the line passing through the points $\left(ct_1, \frac{c}{t_1}\right)$ and $\left(ct_2, \frac{c}{t_2}\right)$ is

Let $S = \{(x,y) \in \mathbb{R}^2 : \frac{y^2}{1+r} - \frac{x^2}{1-r} = 1\}$,where $r \neq \pm 1$. Then $S$ represents

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16x^{2}-9y^{2}+32x+36y-164=0$ and its foci is:

The equation of the conic with focus at $(1, -1)$,directrix along $x - y + 1 = 0$ and with eccentricity $e = \sqrt{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module