If the line y = 2x + lambda is a tangent to t | vedclass

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

Question

(C) The equation of the hyperbola is $36x^2 - 25y^2 = 3600$. Dividing by $3600$, we get $\frac{x^2}{100} - \frac{y^2}{144} = 1$.Here, $a^2 = 100$ and

Vedclass · Accepted Answer

$\pm 16$

Vedclass · Answer

(C) The equation of the hyperbola is $36x^2 - 25y^2 = 3600$. Dividing by $3600$, we get $\frac{x^2}{100} - \frac{y^2}{144} = 1$.Here, $a^2 = 100$ and $b^2 = 144$.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 $c^2 = a^2m^2 - b^2$.Comparing $y = 2x + \lambda$ with $y = mx + c$, we have $m = 2$ and $c = \lambda$.Substituting the values into the condition: $\lambda^2 = (100)(2)^2 - 144$.$\lambda^2 = 400 - 144 = 256$.Therefore, $\lambda = \pm \sqrt{256} = \pm 16$.

If the line $y = 2x + \lambda$ is a tangent to the hyperbola $36x^2 - 25y^2 = 3600$, then $\lambda = $

Similar Questions

The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-

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

For what value of $\gamma$ does the line $y = 2x + \gamma$ touch the hyperbola $16x^{2} - 9y^{2} = 144$?

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

What is the distance between the directrices of the hyperbola $x = 8 \sec \theta, y = 8 \tan \theta$ (in $\sqrt{2}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module