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Ellipse

Section: Mathematics

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Question : 19 of 117

Marks: +1, -0

Let for two distinct values of

p

y=x+p

touch the ellipse

E: \frac{x^2}{4^2} + \frac{y^2}{3^3}=1

A

B

y=x

E

C

D

. Then the area of the quadrilateral

A B C D

[4 Apr 2025 Shift 2]

48
20
36
24
None of Above

Solution:

E: \frac{x^2}{4^2} + \frac{y^2}{3^2}=1

T: y=m x \pm \sqrt{16 m^2+9}

y=x+p

\Rightarrow m=1

\Rightarrow p= \pm \sqrt{16+9}

= \pm 5

T: y=x \pm 5

will to cut the

E

at

A \left(-\frac{16}{5}, \frac{9}{5}\right)

B \left(\frac{16}{5}, -\frac{9}{5}\right)

Also,

y=x

will cut the

E

at

C \left(\frac{12}{5}, \frac{12}{5}\right)

D \left(-\frac{12}{5}, -\frac{12}{5}\right)

A B C D

in not give in cyclic order

\therefore

it does not form any quadrilateral

\therefore

No option should match

If order is not considered then Area

=24

sq. unit.

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