For which of the following curves is the line | vedclass

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(D) The equation of the line is $x + \sqrt{3}y = 2\sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 2$. Here,the slope $m = -\frac{1}{\sqr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of the line is $x + \sqrt{3}y = 2\sqrt{3}$,which can be written as $y = -\frac{1}{\sqrt{3}}x + 2$. Here,the slope $m = -\frac{1}{\sqrt{3}}$.We check the point $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ for each curve.For option $D$: $x^{2} + 9y^{2} = 9$.Substitute the point: $\left(\frac{3\sqrt{3}}{2}\right)^{2} + 9\left(\frac{1}{2}\right)^{2} = \frac{27}{4} + \frac{9}{4} = \frac{36}{4} = 9$. The point lies on the curve.The equation of the tangent to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at $(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}} + \frac{yy_{1}}{b^{2}} = 1$.For $x^{2} + 9y^{2} = 9$,we have $\frac{x^{2}}{9} + y^{2} = 1$.Tangent at $\left(\frac{3\sqrt{3}}{2}, \frac{1}{2}\right)$ is $\frac{x \cdot \frac{3\sqrt{3}}{2}}{9} + y \cdot \frac{1}{2} = 1$.$\frac{\sqrt{3}x}{6} + \frac{y}{2} = 1 \implies \sqrt{3}x + 3y = 6 \implies x + \sqrt{3}y = 2\sqrt{3}$.This matches the given line.

List-$I$	List-$II$
$(P)$ Directrix corresponding to the focus $(-3, 0)$	$(1)$ $y = 4$
$(Q)$ Tangent at the vertex $(0, 4)$	$(2)$ $3x = 25$
$(R)$ Latus rectum through $(3, 0)$	$(3)$ $x = 3$
	$(4)$ $y + 4 = 0$
	$(5)$ $x + 3 = 0$
	$(6)$ $3x + 25 = 0$

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

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