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JEE MATHEMATICS MCQ: TRANSFORMATION OF AXES

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Application +4 | -1

The transformed equation of

{x^2} + 6xy + 8{y^2} = 10

when the axes are rotated through an angle

\frac{\pi }{4}

is

View answer and explanation

Correct option: C

15{x^2} + 14xy + 3{y^2} = 20

Option C (

15{x^2} + 14xy + 3{y^2} = 20

) is correct because after rotating the axes by

\frac{\pi}{4}

, the transformed equation is

15{x^2} + 14xy + 3{y^2} = 20

.

More Questions from TRANSFORMATION OF AXES

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If first degree terms and constant term are to be removed from the equation

12 x^2+7 x y-12 y^2-17 x-31 y-7=0,

then the origin must be shifted at the point The pedal points of a perpendicular drawn from origin on the line

3 x+4 y-5=0

, is In order to remove

xy

-term from the equation

5{x^2} + 4\sqrt 3 \;xy + 9{y^2} - 8 = 0

the coordinate axes must be rotated through an angle For which of the following value of

\alpha=\ldots \ldots

, the slope of the line

x \cos \alpha+y \sin \alpha=p

p \neq 0

\sqrt{3}

. If the origin is shifted to the point

(a b /(a-b), 0)

without rotation, then the equation

(a-b)\left(x^2+y^2\right)-2 a b x-0

becomes - The coordinates of the point where origin should be shifted so that the equation

y^2+4 y+8 x-2=0

will not contain term in

y

and the constant term, are