JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

The locus of chords of contact of perpendicular tangents to the ellipse

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

touch another fixed ellipse is

View answer and explanation

Correct option: C

\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}

Option C (

\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}

) is correct because it represents the equation of the fixed ellipse touched by the chords of contact of perpendicular tangents to the given ellipse.

More Questions from ELLIPSE

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

m

\mathrm{M}

\mathrm{mM}

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

S

S'.

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A

B

(\sqrt{7}, 0)

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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

r

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

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x=0

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R

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y=x

Q

R

N P

Q

P

O