Find the equation of tangents to the ellipse $3 x^{2} + 4 y^{2} = 12$ which are perpendicular to the line $y + 2 x = 4.$ The equation is given by $y = m x \pm b$ , find $m \times b .$

\frac{1}{2}

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Solution

The correct option is B $1$
The ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ $∴ a^{2} = 4, b^{2} = 3$
Slope of the line $y + 2 x = 4$ is $- 2.$
Any tangent to the ellipse is $y = m x \pm \sqrt{(a^{2} m^{2} + b^{2})}$
If it is perpendicular to given line then $m = \frac{1}{2}$
$y = \frac{1}{2} x \pm 2$
$\Rightarrow m \times b = 1$
Ans: B

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Similar questions

Find the equations of the tangents to the ellipse $3 x^{2} + 4 y^{2} = 12$ which are perpendicular to the line $y + 2 x = 4$

3 x^{2} + 4 y^{2} = 12

y = x - 1

3 x^{2} + 4 y^{2} = 12,

y + 2 x = 4,

3 x^{2} + 4 y^{2} = 12,

y + 2 x = 4,

{3 x}^{2} + {4 y}^{2} = 12

y + 2 x = 4

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