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Ellipse

Section: Mathematics

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Question : 63 of 117

Marks: +1, -0

The line

y=x+1

meets the ellipse

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

P

Q

r

is the radius of the circle with

P Q

as diameter then

(3 r)^2

[25-Jun-2022-Shift-2]

20
12
11
8

Solution: 👈: Video Solution

Let point

(a, a+1)

as the point of intersection of line and ellipse.

So,

\;\frac{a^2}{4}+\;\frac{(a+1)^2}{2}=1 \Rightarrow a^2+2(a^2+2a+1)=4

\Rightarrow 3a^2+4a-2=0

If roots of this equation are

\alpha

and

\beta

.

So,

P(\alpha, \alpha+1)

and

Q(\beta, \beta+1)

P Q=4 r^2=(\alpha-\beta)^2+(\alpha-\beta)^2

\Rightarrow 9 r^2=\;\frac{9}{4}(2(\alpha-\beta)^2)

=\;\frac{9}{2}[(\alpha+\beta)^2-4\alpha\beta]

=\;\frac{9}{2}\left[\left(-\;\frac{4}{3}\right)^2+\;\frac{8}{3}\right]

=\;\frac{1}{2}[16+24]=20

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