find radius of curvature at the point (a,0) on the curve y*2=a*2(a-x)/x
Answers
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:
The radius of curvature at the point (a,0) on the curve
The radius of curvature at a point on a curve is given by the formula:
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:
Next, we need to find the first and second derivatives of y(x) with respect to x:
Now we can plug these values into the formula for the radius of curvature:
And we can evaluate this expression at the point (a,0) by replacing x with a:
1
Finally, we can find the radius of curvature by taking the reciprocal of the previous expression:
So the radius of curvature at the point (a,0) on the curve is
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