If frac{x^{2}}{36}-frac{y^{2}}{k^{2}}=1 is a | vedclass

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

Question

(C) Given the equation of the hyperbola: $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$. Rearranging for $k^{2}$: $\frac{y^{2}}{k^{2}} = \frac{x^{2}}{36}-1

Vedclass · Accepted Answer

$(10, 4)$ lies on the hyperbola

Vedclass · Answer

(C) Given the equation of the hyperbola: $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$. Rearranging for $k^{2}$: $\frac{y^{2}}{k^{2}} = \frac{x^{2}}{36}-1 = \frac{x^{2}-36}{36}$ $k^{2} = \frac{36y^{2}}{x^{2}-36}$. Since $k^{2} > 0$,we must have $x^{2}-36 > 0$,which implies $x^{2} > 36$ or $|x| > 6$. Checking the options: $A$. $(-3, 1) \Rightarrow x^{2} = 9 $B$. $(3, 1) \Rightarrow x^{2} = 9 $C$. $(10, 4) \Rightarrow x^{2} = 100 > 36$ (True) $D$. $(5, 2) \Rightarrow x^{2} = 25 Therefore,the point $(10, 4)$ can lie on the hyperbola.

If $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$ is a hyperbola,then which of the following statements can be true?

Similar Questions

At which point does the line $2x + \sqrt{6}y = 2$ touch the hyperbola $x^2 - 2y^2 = 4$?

If the line $y = mx + 7\sqrt{3}$ is normal to the hyperbola $\frac{x^2}{24} - \frac{y^2}{18} = 1$,then a value of $m$ is

What will be the equation of the chord of the hyperbola $25x^2 - 16y^2 = 400$,whose midpoint is $(5, 3)$?

If the equation of the hyperbola having $(8,3)$ and $(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$,then $p+q=$

Find the equation of the tangent to the hyperbola $x^2 - 4y^2 = 36$ which is perpendicular to the line $x - y + 4 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module