A tangent is drawn to the ellipse frac{x^2}{2 | vedclass

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(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a}

Vedclass · Accepted Answer

$\pi/6$

Vedclass · Answer

(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.Here,$a^2 = 27 \implies a = 3\sqrt{3}$ and $b^2 = 1 \implies b = 1$.So,the tangent equation is $\frac{x \cos 	heta}{3\sqrt{3}} + y \sin 	heta = 1$.The $x$-intercept is $X = \frac{3\sqrt{3}}{\cos 	heta} = 3\sqrt{3} \sec 	heta$ and the $y$-intercept is $Y = \frac{1}{\sin 	heta} = \csc 	heta$.Let the sum of intercepts be $f(	heta) = 3\sqrt{3} \sec 	heta + \csc 	heta$.To find the minimum,we set $f'(	heta) = 0$:$f'(	heta) = 3\sqrt{3} \sec 	heta 	an 	heta - \csc 	heta \cot 	heta = 0$.$3\sqrt{3} \frac{\sin 	heta}{\cos^2 	heta} = \frac{\cos 	heta}{\sin^2 	heta}$.$	an^3 	heta = \frac{1}{3\sqrt{3}} = \left(\frac{1}{\sqrt{3}}ight)^3$.$	an 	heta = \frac{1}{\sqrt{3}} \implies 	heta = \frac{\pi}{6}$.

List$-I$	List$-II$
$(P)$ Passing $H _2 S$ in the presence of $NH _4 OH$	$(1) \ Cu ^{2+}$
$(Q) \ \left( NH _4\right)_2 CO _3$ in the presence of $NH _4 OH$	$(2) \ Al ^{3+}$
$(R) \ NH _4 OH$ in the presence of $NH _4 Cl$	$(3) \ Mn ^{2+}$
$(S)$ Passing $H _2 S$ in the presence of dilute $\text{HCl}$	$(4) \ Ba ^{2+}$
	$(5) \ Mg ^{2+}$

$A$ tangent is drawn to the ellipse $\frac{x^2}{27} + y^2 = 1$ at the point $(3\sqrt{3} \cos \theta, \sin \theta)$,where $\theta \in (0, \pi/2)$. The value of $\theta$ for which the sum of the intercepts on the axes made by this tangent is minimum,is:

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