If 4x+y+p=0 (p0) is a tangent to the ellipse | vedclass

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(A) For a line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the condition is $c^2=a^2m^2+b^2$. Given $4x+y+p=0$,we have

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(A) For a line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the condition is $c^2=a^2m^2+b^2$. Given $4x+y+p=0$,we have $y=-4x-p$. For the ellipse $x^2+3y^2=3$,i.e.,$\frac{x^2}{3}+\frac{y^2}{1}=1$,we have $a^2=3, b^2=1, m=-4, c=-p$. Substituting these values: $(-p)^2 = 3(-4)^2 + 1 \Rightarrow p^2 = 48+1 = 49$. Since $p>0$,$p=7$. For a line $y=mx+c$ to be a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the condition is $c^2=\frac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$. Given $16x+qy+14=0$,we have $y=-\frac{16}{q}x-\frac{14}{q}$. For the ellipse $x^2+8y^2=33$,i.e.,$\frac{x^2}{33}+\frac{y^2}{33/8}=1$,we have $a^2=33, b^2=\frac{33}{8}, m=-\frac{16}{q}, c=-\frac{14}{q}$. Substituting these values: $\frac{196}{q^2} = \frac{(-16/q)^2(33-33/8)^2}{33+(33/8)(-16/q)^2} = \frac{(256/q^2)(33 	imes 7/8)^2}{33+(33/8)(256/q^2)} = \frac{(256/q^2)(33^2 	imes 49 / 64)}{33(1+32/q^2)} = \frac{4 	imes 33^2 	imes 49 / q^2}{33(q^2+32)/q^2} = \frac{4 	imes 33 	imes 49}{q^2+32}$. Simplifying: $\frac{196}{q^2} = \frac{6468}{q^2+32}$ $\Rightarrow \frac{49}{q^2} = \frac{1617}{q^2+32}$ $\Rightarrow q^2+32 = 33q^2$ $\Rightarrow 32q^2=32$ $\Rightarrow q^2=1$. Since $q>0$,$q=1$. Therefore,$p+q = 7+1 = 8$.

If $4x+y+p=0$ $(p>0)$ is a tangent to the ellipse $x^2+3y^2=3$ and $16x+qy+14=0$ $(q>0)$ is a normal to the ellipse $x^2+8y^2=33$,then $p+q=$

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