Tangent to ellipse ${3 x}^{2} + {4 y}^{2} = 12$ to $y + 2 x = 4$

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Solution

We have,

Equation of ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$

On comparing that,

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$a^{2} = 4, b^{2} = 3$

Equation of line

$y + 2 x = 4$

$y = - 2 x + 4$

On comparing that,

$y = m x + c$

So,

Slope of this line $m = - 2$

Now equation of a tangent having slope m=-2 is

$y = - 2 x + \sqrt{4 \times 4 + 3}$

$y = - 2 x + \sqrt{19}$

Hence, this is the answer.

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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 + 2 x = 4,

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

2 x - y + 5 = 0

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