JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

The locus of the poles of tangents to the auxiliary circle with respect to the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,

View answer and explanation

Correct option: C

\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2}}}

Option C (

\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} = \frac{1}{{{a^2}}}

) is correct because it represents the locus of the poles of tangents to the auxiliary circle with respect to the given ellipse.

More Questions from ELLIPSE

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\frac{x^2}{144}+\frac{y^2}{81}=1

m

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\mathrm{mM}

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\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

S

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9 x^{2}+16 y^{2}=144

\mathrm{PA}+\mathrm{PB}

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

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\frac{x^2}{25}+\frac{y^2}{16}=1

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x=0

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y=x

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O