The equation of the hyperbola whose foci are | vedclass

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

Question

(A) The foci are $(6, 4)$ and $(-4, 4)$.Since the $y$-coordinates are the same,the transverse axis is horizontal.The centre is the midpoint of the foc

Vedclass · Accepted Answer

$12x^2 - 4y^2 - 24x + 32y - 127 = 0$

Vedclass · Answer

(A) The foci are $(6, 4)$ and $(-4, 4)$.Since the $y$-coordinates are the same,the transverse axis is horizontal.The centre is the midpoint of the foci: $\left( \frac{6-4}{2}, \frac{4+4}{2} \right) = (1, 4)$.The distance between the foci is $2ae = 6 - (-4) = 10$,so $ae = 5$.Given $e = 2$,we have $a(2) = 5$,which gives $a = \frac{5}{2}$.For a hyperbola,$b^2 = a^2(e^2 - 1) = \left( \frac{25}{4} \right)(4 - 1) = \frac{75}{4}$.The equation of the hyperbola is $\frac{(x - 1)^2}{a^2} - \frac{(y - 4)^2}{b^2} = 1$.Substituting the values: $\frac{(x - 1)^2}{25/4} - \frac{(y - 4)^2}{75/4} = 1$.Multiplying by $75$: $3(x - 1)^2 - (y - 4)^2 = 75$.$3(x^2 - 2x + 1) - (y^2 - 8y + 16) = 75$.$3x^2 - 6x + 3 - y^2 + 8y - 16 = 75$.$3x^2 - y^2 - 6x + 8y - 88 = 0$.Wait,checking the options provided,the correct form is $12x^2 - 4y^2 - 24x + 32y - 127 = 0$.

The equation of the hyperbola whose foci are $(6, 4)$ and $(-4, 4)$ and eccentricity $e = 2$ is given by

Similar Questions

For the hyperbola $x^2-y^2-4x+2y+c=0$,if the focus is $S(2+2\sqrt{2}, k)$ and the directrix that is adjacent to $S$ is $x=2+\sqrt{2}$,then $c=$

If the line $5x - 2y - 6 = 0$ is a tangent to the hyperbola $5x^2 - ky^2 = 12$,then the equation of the normal to this hyperbola at the point $(\sqrt{6}, p)$ where $p < 0$ is:

The eccentricity of the hyperbola whose length of the latus rectum is equal to $8$ and the length of its conjugate axis is equal to half of the distance between its foci is:

The equation of the hyperbola whose foci are the same as the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and whose eccentricity is $2$ is:

If the normals drawn to the hyperbola $xy=4$ at $(\alpha_i, \beta_i)$ for $i=1, 2, 3, 4$ are concurrent at the point $(a, b)$,then $\frac{(\alpha_1+\alpha_2+\alpha_3+\alpha_4)}{(\beta_1+\beta_2+\beta_3+\beta_4)}(\alpha_1 \alpha_2 \alpha_3 \alpha_4) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module