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प्रश्न
A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ______.
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उत्तर
A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is 13x2 + 13y2 – 83x + 64y + 172 = 0.
Explanation:
The given equation of line is 5x – 12y = 3 and the given point is (3, – 2).
Let (a, b) be any moving point
∴ Distance between (a, b) and the point (3, – 2)
= `sqrt((a - 3)^2 + (b + 2)^2)`
And the distance of (a, b) from the line 5x – 12y = 3
= `|(5a - 12b - 3)/sqrt(25 + 144)|`
= `|(5a - 12b - 3)/13|`
We have `[sqrt((a - 3)^2 + (b + 2)^2)]^2 = |(5a - 12b - 3)/13|`
⇒ `(a - 3)^2 + (b + 2)^2 = (5a - 12b - 3)/13`
Taking numerical values only, we have
`(a - 3)^2 + (b + 2)^2 = (5a - 12b - 3)/13`
⇒ `a^2 - 6a + 9 + b^2 + 4b + 4 = (5a - 12b - 3)/13`
⇒ `a^2 + b^2 - 6a + 4b + 13 = (5a - 12b - 3)/13`
⇒ 13a2 + 13b2 – 78a + 52b + 169 = 5a – 12b – 3
⇒ 13a2 + 13b2 – 83 + 64b + 172 = 0
So, the locus of the point is 13x2 + 13y2 – 83x + 64y + 172 = 0.
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