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Standard XII

Mathematics

Formation of a Differential Equation from a General Solution

Find what the...

Question

Find what the following equation $x^{2} - y^{2} + 2 x + y = 0$ becomes when the origin is shifted to the point (1, 1)

x^{2} - y^{2} + 4 x - y + 3 = 0

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x y - y^{2} - x + y = 0

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x y - x - y + 1 = 0

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None of these

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Solution

The correct option is B $x^{2} - y^{2} + 4 x - y + 3 = 0$
When the origin is shifted to the point $(1, 1)$ then $x = x^{'} + 1, y = y^{'} + 1$

Substituting the above value in the given equation, we get

${(x^{'} + 1)}^{2} - {(y^{'} + 1)}^{2} + 2 (x^{'} + 1) + y^{'} + 1 = 0$

$\Rightarrow {x^{'}}^{2} + 1 + 2 x^{'} - {y^{'}}^{2} - 1 - 2 y^{'} + 2 x^{'} + 2 + y^{'} + 1 = 0$

$\Rightarrow {x^{'}}^{2} - {y^{'}}^{2} + 4 x^{'} - y^{'} + 3 = 0$

Replacing $x^{'} \to x$ and $y^{'} \to y$ we get

$\Rightarrow x^{2} - y^{2} + 4 x - y + 3 = 0$

Thus, the new equation of the curve is $x^{2} - y^{2} + 4 x - y + 3 = 0$

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Similar questions

Q. Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x² + xy − 3y² − y + 2 = 0
(ii) xy − y² − x + y = 0
(iii) xy − x − y + 1 = 0
(iv) x² − y² − 2x + 2y = 0

Q. Find what the following equations become when the origin is shifted to the point

(1, 1)

?
(i)

x^{2} + x y - 3 x - y + 2 = 0

(ii)

x^{2} - y^{2} - 2 x + 2 y = 0

(iii)

x y - x - y + 1 = 0

(iv)

x y - y^{2} - x + y = 0

$Find what the following equations become when the origin is shifted to the point(1,1) ? (i) x^{2} + x y - 3 y - y + 2 = 0 (i i) x^{2} - y^{2} - 2 x + 2 y = 0 (i i i) x y - x - y + 1 = 0 (i v) x y - y^{2} - x + y = 0$

Q. If the origin is shifted to the point (1, 1), axes remaining parallel, find the new equation of the locus in each of the following.
i)

x y - x - y + 1 = 0

ii)

x^{2} - y^{2} - 2 x + 2 y = 0

iii)

x^{2} + y^{2} - 4 x + 6 y + 3 = 0

Q. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = e^{x} + 1

y^{''} - y = 0

y = x^{2} + 2 x + C

y^{'} - 2 x - 2 = 0

y = cos x + C

y^{'} + sin x = 0

y = \sqrt{1 + x^{2}}

y^{'} = \frac{x y}{1 + x^{2}}

y = A x

x y = y (x \neq 0)

y = x sin x

x y = y + x \sqrt{x^{2} y^{2}} (x \neq 0 a n d x > y o r x < y)

x y = log y + C

y^{'} = \frac{y^{2}}{1 - x y} (x y \neq 1)

y - cos y = x

(y sin y + cos y + x) y^{'} = y

x + y = {tan}^{- 1} y

y^{2} y^{'} + y^{2} + 1 = 0

10.

y = \sqrt{a^{2} - x^{2}} x ϵ (- a, a)

x + y \frac{d y}{d x} = 0 (y \neq 0)

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Find what the following equation x2−y2+2x+y=0 becomes when the origin is shifted to the point (1, 1)

Find what the following equation $x^{2} - y^{2} + 2 x + y = 0$ becomes when the origin is shifted to the point (1, 1)