JEE MATHEMATICS MCQ: ELLIPSE

1 Easy

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Application +4 | -1

The locus of the point of intersection of the perpendicular tangents to the ellipse

\frac{x^2}{9}+\frac{y^2}{4}=1

View answer and explanation

Correct option: C

x^2+y^2=13

Option C (

x^2+y^2=13

) is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is the director circle, given by

x^2 + y^2 = a^2 + b^2

More Questions from ELLIPSE

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

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\mathrm{mM}

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

S

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A

B

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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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y=x

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O