The equation of a conic section whose focus is at $(- 1, 0)$ , directrix is the line $4 x - 3 y + 2 = 0$ and eccentricity is $\frac{1}{\sqrt{2}}$ , is

34 x^{2} + 24 x y + 41 y^{2} + 84 x + 12 y + 46 = 0

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

41 x^{2} + 24 x y + 34 y^{2} + 84 x + 12 y + 46 = 0

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

34 x^{2} + 24 x y + 41 y^{2} + 12 x + 84 y + 46 = 0

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

34 x^{2} + 24 x y + 41 y^{2} + 84 x + 12 y + 81 = 0

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

Open in App

Solution

The correct option is A $34 x^{2} + 24 x y + 41 y^{2} + 84 x + 12 y + 46 = 0$
$e = \frac{1}{\sqrt{2}}$ $4 x - 3 y + 2 = 0$ directrix focus $= (- 1, 0)$
$f = \sqrt{(x + 1)^{2} + (y - 0)^{2}}$
distance from $(x, y)$ to the line $4 x - 3 y + 2 = 0$
$d = ∣ ∣ ∣ \frac{4 x - 3 y + 2}{\sqrt{25}} ∣ ∣ ∣ = ∣ ∣ ∣ \frac{4 x - 3 y + 2}{5} ∣ ∣ ∣$
$e = \frac{f}{d} \Rightarrow \frac{1}{\sqrt{2}} = \frac{5 \sqrt{(x + 1)^{2} + (y - 0)^{2}}}{(4 x - 3 y + 2)^{2}}$
$\frac{1}{2} = \frac{5 (x^{2} + 2 x + 1 + y^{2})}{(4 x - 3 y)^{2} + 4 + 16 \times - 12 y}$
$16 x^{2} + 9 y^{2} - 14 x y + 4 + 16 x - 12 y = 50 x^{2} + 100 x + 50 + 50$
$34 x^{2} + 41 y^{2} + 24 x y + 34 x + 12 y + 46 = 0$

Suggest Corrections

Similar questions

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

\frac{5}{4}

(a, 0)

4 x - 3 y = a

7 y^{2} + 24 x y - 24 a x - 16 a y + 15 a^{2} = 0

3 x + 4 y - 5 = 0

(1, 2)

\frac{1}{2}

91 x^{2} + 84 y^{2} - 24 x y - 170 x - 360 y + 475 = 0

4 x^{2} + 24 x y + 11 y^{2} = 0

3 x + 12 y - 19 = 0

9 x + 3 y - 46 = 0

Join BYJU'S Learning Program