Show that the area of the triangle formed by the tangent and the normal at tha point(a,a) on tha curve y^{2}(2a-x)=x^{3} and the line x=2a,is 5a^{2/}4.

Equation of the given curve is *y*^{2} (2*a* – *x*) = *x*^{3}

*y*^{2} (2*a* – *x*) = *x*^{3}

Differentiating both sides w.r.t *x*, we get

Slope of tangent at

∴ Slope of normal at

Equation of tangent at (*a*, *a*) is

(*y* – *a*) = 2 (*x* – *a*)

∴ *y* = 2*x* – 2*a* +* a *= 2*x* – *a* ...(1)

Equation of normal at (*a*, *a*) is

∴ 2*y* – 2*a* = – *x* + *a*

Equation of the given line is

*x* = 2*a* ...(3)

Solving (1) and (3), we have

*x* = 2*a* and *y* = 2 (2*a*) – *a* = 4*a* – *a* = 3*a*

Solving (2) and (3), we have

Solving (1) and (2), we have

*x* = *a* and *y* = *a*

be the vertices of the triangle.

Are of ∆ABC

Thus, the area of triangle formed is square units.

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