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

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Question

(b) We know that equation of polar at point $(h,\,k)$ is $\frac{{hx}}{{{a^2}}} + \frac{{ky}}{{{b^2}}} = 1$

$ \Rightarrow $ $\frac{{hx}}{4} + \frac{

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(b) We know that equation of polar at point $(h,\,k)$ is $\frac{{hx}}{{{a^2}}} + \frac{{ky}}{{{b^2}}} = 1$

$ \Rightarrow $ $\frac{{hx}}{4} + \frac{{ky}}{1} = 1$

$ \Rightarrow hx + 4ky = 4$ .....$(i)$

Which is similar to given straight line $x + 4y = 4$ &hellip;..$(ii)$

Comparing $(i)$ and $(ii),$ we get $h = 1,\,k = 1$.

Hence the point is $(1,\,1).$

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

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