Assertion (A): The image of frac{x^2}{25}+fra | vedclass

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(D) The given ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$ with center $(0,0)$.To find the image of the ellipse in the line $x+y-10=0$,we find the ima

Vedclass · Accepted Answer

$(A)$ is false but $(R)$ is true

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(D) The given ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$ with center $(0,0)$.To find the image of the ellipse in the line $x+y-10=0$,we find the image of the center $(0,0)$ in the line $x+y-10=0$.Let the image be $(h,k)$. The line joining $(0,0)$ and $(h,k)$ is perpendicular to $x+y-10=0$,so its slope is $1$. Thus,$\frac{k-0}{h-0} = 1 \implies k=h$.The midpoint $(\frac{h}{2}, \frac{k}{2})$ lies on $x+y-10=0$,so $\frac{h}{2}+\frac{k}{2}=10 \implies h+k=20$.Substituting $k=h$,we get $2h=20 \implies h=10, k=10$.The image of the center is $(10,10)$.The shape and size of the ellipse remain unchanged,so the new equation is $\frac{(x-10)^2}{25}+\frac{(y-10)^2}{16}=1$.Comparing this with the given equation $\frac{(x-10)^2}{16}+\frac{(y-10)^2}{25}=1$,we see the denominators are swapped.Thus,Assertion $(A)$ is false.Reason $(R)$ is a standard definition of the image of a curve,which is true.Therefore,$(A)$ is false but $(R)$ is true.

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