Pole of straight line x + 4y = 4 with respect to ellipse {x^2} + 4{y^2} = 4 is

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10-2. Parabola, Ellipse, Hyperbola

hard

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

A

$(1, 4)$

B

$(1, 1)$

C

$(4, 1)$

D

$(4, 4)$

Solution

(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 $x + 4y = 4$ …..$(ii)$

Comparing $(i)$ and $(ii),$ we get $h = 1,\,k = 1$.

Hence the point is $(1,\,1).$

Standard 11

Mathematics

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