Math, asked by pandulababylikhitha, 1 year ago

find radius of curvature at the point (a,0) on the curve y*2=a*2(a-x)/x​

Answers

Answered by AdityaRohan
3

Answer:

The equation of the curve, (a – x)y2 = (a + x)x2 passes through the origin. To see the nature of the tangent at the origin, equate to zero the lowest degree terms in x and y, i.e. ay2 = ax2 or y = ± x i.e., at the origin, neither of the axis are tangent to the given curve ∴ Putting or On comparing the coefficients of x2 and x3, we get ap2 = a ⇒ p = ± 1 and apq – p2 = 1 ⇒ q = ± 2/ a ∴ Hence ρ(0, 0) is numerically a√2.

Step-by-step explanation:

Answered by Afreenakbar
0

The radius of curvature at the point (a,0) on the curve y^2 = a^2(a-x)/x is a^2

The radius of curvature at a point on a curve is given by the formula:

1/R = (y''(x)) / (1 + y'(x)^2)^(3/2)

where R is the radius of curvature and y(x) is the equation of the curve.

To find the radius of curvature at the point (a, 0) on the curve y^2 = a^2(a-x)/x, we first need to find the equation of the curve in the form y(x). We can do this by taking the square root of both sides:

y = +/- \sqrt(a^2(a-x)/x)

Next, we need to find the first and second derivatives of y(x) with respect to x:

y'(x) = -a^2/x^2 + a/x

y''(x) = 2a^2/x^3 - a/x^2

Now we can plug these values into the formula for the radius of curvature:

1/R = (2a^2/x^3 - a/x^2) / (1 + (-a^2/x^2 + a/x)^2)^(3/2)

And we can evaluate this expression at the point (a,0) by replacing x with a:

1/R = (2a^2/a^3 - a/a^2) / (1 + (-a^2/a^2 + a/a)^2)^(3/2)

1/R = (-a/a^2) / (1 + (a^2/a^2)^2)^(3/2)

1/R = -1/a^2

Finally, we can find the radius of curvature by taking the reciprocal of the previous expression:

R = a^2

So the radius of curvature at the point (a,0) on the curve y^2 = a^2(a-x)/xis a^2

To know more about  radius visit : https://brainly.in/question/23536853

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