For which of the following curves- the line x-3 y-2 -3 is the tangent at the point 3 -3-2- 1-2 -

The correct option is A x^2+9y^2=9Tangent to x^2+9y^2=9 at point ( nbsp; 3√(3)2,12 ) is T=0 i.e., nbsp;x (3√(3)2 nbsp; )+9y ( 12 ) 9=0 ⇒ 3√(3)x+9y=18 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Point Form of Tangent: Ellipse

For which of ...

Question

For which of the following curves, the line $x + \sqrt{3} y = 2 \sqrt{3}$ is the tangent at the point $(\frac{3 \sqrt{3}}{2}, \frac{1}{2})$ ?

x^{2} + 9 y^{2} = 9

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2 x^{2} - 18 y^{2} = 9

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y^{2} = \frac{1}{6 \sqrt{3}} x

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x^{2} + y^{2} = 7

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Solution

The correct option is A $x^{2} + 9 y^{2} = 9$
Tangent to $x^{2} + 9 y^{2} = 9$ at point $(\frac{3 \sqrt{3}}{2}, \frac{1}{2})$ is $T = 0$
i.e., $x (\frac{3 \sqrt{3}}{2}) + 9 y (\frac{1}{2}) - 9 = 0$
$\Rightarrow 3 \sqrt{3} x + 9 y = 18$
$\Rightarrow x + \sqrt{3} y = 2 \sqrt{3}$

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For which of the following curves, the line x+√3y=2√3 is the tangent at the point (3√32,12) ?

The correct option is A x2+9y2=9Tangent to x2+9y2=9 at point (3√32,12) is T=0 i.e., x(3√32)+9y(12)−9=0 ⇒3√3x+9y=18 ⇒x+√3y=2√3

For which of the following curves, the line $x + \sqrt{3} y = 2 \sqrt{3}$ is the tangent at the point $(\frac{3 \sqrt{3}}{2}, \frac{1}{2})$ ?

The correct option is A $x^{2} + 9 y^{2} = 9$
Tangent to $x^{2} + 9 y^{2} = 9$ at point $(\frac{3 \sqrt{3}}{2}, \frac{1}{2})$ is $T = 0$
i.e., $x (\frac{3 \sqrt{3}}{2}) + 9 y (\frac{1}{2}) - 9 = 0$
$\Rightarrow 3 \sqrt{3} x + 9 y = 18$
$\Rightarrow x + \sqrt{3} y = 2 \sqrt{3}$