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

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Solution

Slope of tangent $= m = \frac{1}{2}$ Given
ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ Equation
of tangent whose slope is m is $y = m x \pm \sqrt{4 m^{2} + 3} ∵ m = \frac{1}{2} ∴ y = \frac{1}{2} x \pm \sqrt{1 + 3} 2 y = x \pm 4$

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Similar questions

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

y + 2 x = 4.

y = m x \pm b

m \times b .

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

x^{2} - 4 y^{2} = 36

x - y + 4 = 0

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