If the vertices of a hyperbola are at (-2, 0) | vedclass

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

Question

(D) The standard equation of a hyperbola with vertices at $(\pm a, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given vertices are at $(\pm 2, 0)$,s

Vedclass · Accepted Answer

$(6, 5\sqrt{2})$

Vedclass · Answer

(D) The standard equation of a hyperbola with vertices at $(\pm a, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given vertices are at $(\pm 2, 0)$,so $a = 2$,which implies $a^2 = 4$.The focus is at $(\pm ae, 0)$. Given one focus is at $(-3, 0)$,we have $ae = 3$.For a hyperbola,the relationship between $a, b,$ and $e$ is $b^2 = a^2(e^2 - 1) = a^2e^2 - a^2$.Substituting the values,$b^2 = 3^2 - 2^2 = 9 - 4 = 5$.Thus,the equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{5} = 1$.Now,we check which point does not satisfy this equation:For option $A$: $\frac{(-6)^2}{4} - \frac{(2\sqrt{10})^2}{5} = \frac{36}{4} - \frac{40}{5} = 9 - 8 = 1$ (Lies on hyperbola).For option $B$: $\frac{(2\sqrt{6})^2}{4} - \frac{5^2}{5} = \frac{24}{4} - \frac{25}{5} = 6 - 5 = 1$ (Lies on hyperbola).For option $C$: $\frac{4^2}{4} - \frac{(\sqrt{15})^2}{5} = \frac{16}{4} - \frac{15}{5} = 4 - 3 = 1$ (Lies on hyperbola).For option $D$: $\frac{6^2}{4} - \frac{(5\sqrt{2})^2}{5} = \frac{36}{4} - \frac{50}{5} = 9 - 10 = -1 
eq 1$ (Does not lie on hyperbola).Therefore,the point $(6, 5\sqrt{2})$ does not lie on the hyperbola.

If the vertices of a hyperbola are at $(-2, 0)$ and $(2, 0)$ and one of its foci is at $(-3, 0)$,then which one of the following points does not lie on this hyperbola?

Similar Questions

If $P(\theta)$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $S$ and $S^{\prime}$ are the foci of the hyperbola,then $SP \cdot S^{\prime}P =$

If $PQ$ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that $\Delta OPQ$ is equilateral, where $O$ is the centre, then the eccentricity $e$ satisfies:

The equations of the tangents to the hyperbola $4x^2 - y^2 = 12$ are $y = 4x + c_1$ and $y = 4x + c_2$. Then $|c_1 - c_2|$ is equal to -

The radius of the director circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$,then the coordinates of the point of contact are

Vedclass Test Series

Exam Paper Generator

Online Exam Module