Advertisements
Advertisements
प्रश्न
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.
Advertisements
उत्तर

It is given that the distance between the foot of ordinate of point of contact (A) and point of intersection of tangent with x-axis (T) = 2x
\[\text{Coordinate of }T = \left( x - y\frac{dx}{dy}, 0 \right)\]
\[\Rightarrow y - 0 = \frac{dy}{dx}\left(x - \left( x - y \frac{dx}{dy} \right) \right)\]
\[\Rightarrow y \frac{dx}{dy} = 2x\]
\[\Rightarrow \int\frac{dx}{x} = 2\int\frac{dy}{y}\]
\[\Rightarrow \ln x = \ln y^2 + \ln c\]
\[x = c y^2 \]
\[\text{As the circle passes through }\left( 1, 2 \right).\]
\[1 = c \times 2^2 \]
\[ \Rightarrow c = \frac{1}{4}\]
\[ \Rightarrow 4x = y^2\]
APPEARS IN
संबंधित प्रश्न
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} = y\]
|
y = ax |
x cos y dy = (xex log x + ex) dx
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The differential equation satisfied by ax2 + by2 = 1 is
Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
y = ex + 1 y'' − y' = 0
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
The solution of `dy/ dx` = 1 is ______.
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
y2 dx + (xy + x2)dy = 0
x2y dx – (x3 + y3) dy = 0
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
Solve the differential equation
`y (dy)/(dx) + x` = 0
The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.
