JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

The locus of point of intersection of perpendicular tangents of ellipse

\frac{(x-1)^2}{16}+\frac{(y-1)^2}{9}=1

is -

View answer and explanation

Correct option: C

x^2+y^2-2 x-2 y-23=0

Option C (

x^2+y^2-2 x-2 y-23=0

) is correct because the locus of perpendicular tangents of an ellipse is its director circle, which is found by

(x-h)^2 + (y-k)^2 = a^2 + b^2

More Questions from ELLIPSE

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

m

\mathrm{M}

\mathrm{mM}

P

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

S

S'.

A

PSS',

A

A

B

(\sqrt{7}, 0)

(-\sqrt{7}, 0)

P

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

P

x=0

Q

R

Q

y=x

Q

R

N P

Q

P

O