When is the line ℓx + my + n = 0 a tangent to | vedclass

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

Question

(A) The line $y = mx + c$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if and only if $c^2 = a^2m^2 - b^2$.For the given line

Vedclass · Accepted Answer

$a^2ℓ^2 - b^2m^2 = n^2$

Vedclass · Answer

(A) The line $y = mx + c$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if and only if $c^2 = a^2m^2 - b^2$.For the given line $ℓx + my + n = 0$,we can rewrite it as $y = -(\frac{ℓ}{m})x - (\frac{n}{m})$.Here,the slope is $M = -\frac{ℓ}{m}$ and the intercept is $C = -\frac{n}{m}$.Substituting these into the condition $C^2 = a^2M^2 - b^2$,we get:$(-\frac{n}{m})^2 = a^2(-\frac{ℓ}{m})^2 - b^2$$\frac{n^2}{m^2} = \frac{a^2ℓ^2}{m^2} - b^2$Multiplying both sides by $m^2$,we obtain $n^2 = a^2ℓ^2 - b^2m^2$,which can be rearranged as $a^2ℓ^2 - b^2m^2 = n^2$.

When is the line $ℓx + my + n = 0$ a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$?

Similar Questions

The equation $\frac{x^2}{29 - p} + \frac{y^2}{4 - p} = 1$ $(p \neq 4, 29)$ represents:

The point of contact of the tangent $y = x + 2$ to the hyperbola $5x^2 - 9y^2 = 45$ is

If the equation of a hyperbola is $9x^2 - 16y^2 + 72x - 32y - 16 = 0$,then the equation of its conjugate hyperbola is

The $X$ and $Y$ intercepts of the tangent to the hyperbola $\frac{x^2}{20}-\frac{y^2}{5}=1$ which is perpendicular to the line $4x+3y=7$,are respectively

If a hyperbola has asymptotes $3x - 4y - 1 = 0$ and $4x - 3y - 6 = 0$,then the transverse and conjugate axes of that hyperbola are

Vedclass Test Series

Exam Paper Generator

Online Exam Module