Equation of the hyperbola with focus (-3,4) directrix 3x-4y+5=0 and e = $\frac{5}{2}$ is

$5 x^{2} - 24 x y + 12 y^{2} + 6 x - 8 y - 75 = 0$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$5 x^{2} - 24 x y + 12 y^{2} - 8 x - 6 y - 25 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$5 x^{2} - 24 x y + 12 y^{2} - 12 x + 8 y - 55 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$5 x^{2} - 24 x y + 12 y^{2} - 7 x - 12 y - 65 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$5 x^{2} - 24 x y + 12 y^{2} + 6 x - 8 y - 75 = 0$

S = ( -3,4) 3x - 4y + 5 = 0 $e = \frac{5}{2}$

$(x + 3)^{2} + (y - 4)^{2} = \frac{25}{4} \frac{(3 x - 4 y + 5)^{2}}{25}$

$\Rightarrow$ $4 x^{2} + 4 y^{2} + 24 x - 32 y + 100 = 9 x^{2} + 16 y^{2} + 25 - 24 x y + 30 x - 40 y$

$\Rightarrow$ $5 x^{2} - 24 x y + 12 y^{2} + 6 x - 8 y - 75 = 0$

Suggest Corrections

Similar questions

Equation of the hyperbola with focus (-3,4) directrix 3x-4y+5=0 and e = $\frac{5}{2}$ is

Q. The equation of a conic with directrix

3 x + 4 y - 5 = 0

and focus (0, 0) is given as

x^{2} + y^{2} = (3 x + 4 y - 5)^{2}

. Find eccentricity of the conic.

Find the equation of the ellipse in the following cases:
(i) focus is (0,1), directrix is x+y=0 and $e = \frac{1}{2}$ .
(ii) focus is (-1,1), directrix is x-y+3=0 and $e = \frac{1}{2}$ .
(iii) focus is (-2,3), directrix is 2x+3y+4=0 and $e = \frac{4}{5}$ .
(iv) focus is (1,2), directrix is 3x+4y-5=0 and $e = \frac{1}{2}$ .