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

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(B) 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 $b^2 = 1$.Let the p

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$(1, 1)$

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(B) 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 $b^2 = 1$.Let the pole be $(h, k)$. The equation of the polar of the point $(h, k)$ with respect to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{hx}{a^2} + \frac{ky}{b^2} = 1$.Substituting the values,we get $\frac{hx}{4} + \frac{ky}{1} = 1$,which simplifies to $hx + 4ky = 4$.Comparing this with the given line $x + 4y = 4$,we get $h = 1$ and $k = 1$.Therefore,the pole is $(1, 1)$.

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

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