The equation of the tangents to the ellipse $4 x^{2} + 3 y^{2} = 5$ which are parallel to the line $y = 3 x + 7$ are

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Solution

Given ellipse $4 x^{2} + 3 y^{2} = 5 \frac{4 x^{2}}{5} + \frac{3 y^{2}}{5} = 1 \Rightarrow \frac{x^{2}}{\frac{5}{4}} + \frac{y^{2}}{\frac{5}{3}} = 1$ here $a^{2} = \frac{5}{3} b^{2} = \frac{5}{4}$
Given line $y = 3 x + 7$ Slope $= 3$
Equation of tangents $y = m x \pm \sqrt{a^{2} + b^{2} m^{2}} y = 3 x \pm \sqrt{\frac{5}{3} + \frac{5 \times 9}{4}} y = 3 x \pm \sqrt{\frac{155}{12}} \Rightarrow 2 \sqrt{3} y = 6 \sqrt{3} x \pm 155$

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Similar questions

4 x^{2} + 3 y^{2} = 5

3 x - y + 7 = 0

4 x^{2} + 3 y^{2} = 5

y = 3 x + 7

4 x^{2} + 3 y^{2} = 5

y = 3 x + 7

60^{o}

x

2 x^{2} - 3 y^{2} = 6

y = 3 x + 4

x^{2} - 3 y^{2} = 9

y = 3 x + 5.

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