Equation of the two tangents drawn from $(2, - 1)$ to $x^{2} + 3 y^{2} = 3$ are:

x = 2, 4 x + 3 y - 7 = 0

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

y = - 1, 4 x + y - 7 = 0

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

x + y - 1 = 0, 4 x + y - 7 = 0

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

x + y + 1 = 0, 4 x + y - 7 = 0

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

Open in App

Solution

The correct option is B $y = - 1, 4 x + y - 7 = 0$
Line passing through $(2, - 1)$ is

$y + 1 = m (x - 2)$

$y = m x + (- 1 - 2 m)$

Condition of tangency is $c^{2} = a^{2} m^{2} + b^{2}$

$(- 1 - 2 m)^{2} = 3 m^{2} + 1$

$1 + 4 m^{2} + 4 m = 3 m^{2} + 1$

$m^{2} + 4 m = 0 \Rightarrow m = 0, - 4$

$∴$ equation of tangents are $y = - 1$ and $4 x + y = 7$

Suggest Corrections

Similar questions

x + y - 4 = 0, x - y + 2 = 0, x + y = 3

2

2 x + 3 y - 7 = 0, 4 x + 6 y - 1 = 0, x + y + 1 = 0

3

x^{2} + y^{2}

x = 7

(1, - 1),

x - y - 3 = 0

\frac{1}{2}

(x + y) + λ (2 x - y + 1) = 0

(1, - 3)