If x+y+n=0, n0 is a normal to the ellipse | vedclass

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(B) The equation of the ellipse is $x^2+3y^2=3$,which can be written as $\frac{x^2}{3}+y^2=1$. Here $a^2=3$ and $b^2=1$.
The condition for $lx+my+n=0

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$x-5y+5=0$

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(B) The equation of the ellipse is $x^2+3y^2=3$,which can be written as $\frac{x^2}{3}+y^2=1$. Here $a^2=3$ and $b^2=1$.
The condition for $lx+my+n=0$ to be a normal to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2)^2}{n^2}$.
Substituting $l=1, m=1, a^2=3, b^2=1$,we get $\frac{3}{1^2}+\frac{1}{1^2}=\frac{(3-1)^2}{n^2} \Rightarrow 4=\frac{4}{n^2} \Rightarrow n^2=1$.
Since $n>0$,$n=1$. The normal is $x+y+1=0$.
The equation of the ellipse is $x^2+5y^2=5$,or $\frac{x^2}{5}+y^2=1$.
The condition for $lx+my+n=0$ to be a tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $n^2=a^2l^2+b^2m^2$.
Substituting $l=1, n=3, a^2=5, b^2=1$,we get $3^2=5(1)^2+1(m)^2 \Rightarrow 9=5+m^2 \Rightarrow m^2=4$.
Since $m < 0$,$m=-2$. The tangent is $x-2y+3=0$.
Solving $x+y+1=0$ and $x-2y+3=0$: Subtracting the equations gives $3y-2=0 \Rightarrow y=\frac{2}{3}$.
Then $x=-1-y=-1-\frac{2}{3}=-\frac{5}{3}$.
Checking the point $(-\frac{5}{3}, \frac{2}{3})$ in the options: For $x-5y+5=0$,we have $-\frac{5}{3}-5(\frac{2}{3})+5 = \frac{-5-10+15}{3} = 0$.
Thus,the point satisfies $x-5y+5=0$.

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 $(3, 0)$	$(3)$ $x = 3$
	$(4)$ $y + 4 = 0$
	$(5)$ $x + 3 = 0$
	$(6)$ $3x + 25 = 0$

If $x+y+n=0, n>0$ is a normal to the ellipse $x^2+3y^2=3$ and $x+my+3=0, m < 0$ is a tangent to the ellipse $x^2+5y^2=5$,then the point of intersection of these two lines satisfies the equation

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