The line y = mx + c touches the curve frac{x^ | 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 derived by substituting $y = m

Vedclass · Accepted Answer

$c^2 = a^2m^2 - b^2$

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 derived by substituting $y = mx + c$ into the equation of the hyperbola.Substituting $y$ gives $\frac{x^2}{a^2} - \frac{(mx + c)^2}{b^2} = 1$.Multiplying by $a^2b^2$,we get $b^2x^2 - a^2(m^2x^2 + 2mcx + c^2) = a^2b^2$.Rearranging as a quadratic in $x$: $(b^2 - a^2m^2)x^2 - (2a^2mc)x - (a^2c^2 + a^2b^2) = 0$.Since the line is a tangent,the discriminant $D = 0$.$D = (2a^2mc)^2 - 4(b^2 - a^2m^2)(-a^2c^2 - a^2b^2) = 0$.Simplifying this leads to the condition $c^2 = a^2m^2 - b^2$.

The line $y = mx + c$ touches the curve $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,if

Similar Questions

The eccentricity of the conic ${x^2} - 4{y^2} = 1$ is

The equation of the hyperbola whose coordinates of the foci are $(\pm 8, 0)$ and the length of the latus rectum is $24$ units,is

If two tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that the product of their gradients is $c^2$,then they intersect on the curve:

If a hyperbola has length of its conjugate axis equal to $5$ and the distance between its foci is $13$,then the eccentricity of the hyperbola is

The line $lx + my + n = 0$ will be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module