Math, asked by aayza9490, 12 hours ago

The equation of the locus of a point whichmoves so that the sum of its distances from(3,0) and (-3,0) is lessthan 9 is​

Answers

Answered by shadowsabers03
5

Let P(x, y) be our point. Its distance from the point (-3, 0) is,

  • \small\text{$d_1=\sqrt{(x+3)^2+y^2}=\sqrt{x^2+y^2+6x+9}$}

and its distance from the point (3, 0) is,

  • \small\text{$d_2=\sqrt{(x-3)^2+y^2}=\sqrt{x^2+y^2-6x+9}$}

The sum of these two distances is less than 9, i.e.,

\small\text{$\longrightarrow d_1+d_2<9$}

\small\text{$\longrightarrow\sqrt{x^2+y^2+6x+9}+\sqrt{x^2+y^2-6x+9}<9$}

\small\text{$\longrightarrow\sqrt{x^2+y^2+6x+9}<9-\sqrt{x^2+y^2-6x+9}$}

Squaring both sides,

\small\text{$\longrightarrow x^2+y^2+6x+9<\big(9-\sqrt{x^2+y^2-6x+9}\big)^2$}

\small\text{$\longrightarrow x^2+y^2+6x+9<81+x^2+y^2-6x+9-18\sqrt{x^2+y^2-6x+9}$}

\small\text{$\longrightarrow6x<81-6x-18\sqrt{x^2+y^2-6x+9}$}

\small\text{$\longrightarrow18\sqrt{x^2+y^2-6x+9}<81-12x$}

\small\text{$\longrightarrow\sqrt{x^2+y^2-6x+9}<\dfrac{81}{18}-\dfrac{12}{18}x$}

\small\text{$\longrightarrow\sqrt{x^2+y^2-6x+9}<\dfrac{9}{2}-\dfrac{2}{3}x$}

Squaring both sides,

\small\text{$\longrightarrow x^2+y^2-6x+9<\left(\dfrac{9}{2}-\dfrac{2}{3}x\right)^2$}

\small\text{$\longrightarrow x^2+y^2-6x+9<\dfrac{81}{4}-6x+\dfrac{4}{9}x^2$}

\small\text{$\longrightarrow x^2\left(1-\dfrac{4}{9}\right)+y^2<\dfrac{81}{4}-9$}

\small\text{$\longrightarrow\dfrac{5}{9}x^2+y^2<\dfrac{45}{4}$}

\small\text{$\longrightarrow\underline{\underline{\dfrac{4}{81}x^2+\dfrac{4}{45}y^2<1}}$}


amansharma264: Excellent
Similar questions