The pole of the straight line x+4y=4 with res | vedclass

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(A) The equation of the line is $x+4y-4=0$,which is in the form $lx+my+n=0$,where $l=1, m=4, n=-4$. The equation of the ellipse is $x^2+4y^2=4$,which

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$(1,1)$

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(A) The equation of the line is $x+4y-4=0$,which is in the form $lx+my+n=0$,where $l=1, m=4, n=-4$. The equation of the ellipse is $x^2+4y^2=4$,which can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$. Here,$a^2=4$ and $b^2=1$. The formula for the pole $(x_1, y_1)$ of the line $lx+my+n=0$ with respect to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by: $x_1 = \frac{-a^2l}{n}$ and $y_1 = \frac{-b^2m}{n}$. Substituting the values: $x_1 = \frac{-(4)(1)}{-4} = 1$ $y_1 = \frac{-(1)(4)}{-4} = 1$ Thus,the pole is $(1,1)$.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

The pole of the straight line $x+4y=4$ with respect to the ellipse $x^2+4y^2=4$ is

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Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

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