AP EAMCET202418 May 2024Morning ShiftMathematicsApplication of DerivativesActual

The length of the tangent drawn at the point p ( 4 ) on the curve x^ 2 3 +y^ 2 3 =2^ 2 3 is

Options

A2 3
B1
C4 3
D2

Correct answer

B. 1

Step-by-step solution

Given the curve x^ 2 3 +y^ 2 3 =2^ 2 3 Let x=2 ^3 , y=2 ^3 Now, d y d x =- at = 4 , d y d x =-1 and x= 2 2 2 = 1 2 and, y= 1 2 Now, equation of tangent at ( 1 2 , 1 2 ) is y- 1 2 =-1 (x- 1 2 ) y+x= 2 If y=0 x= 2 So, required length = ( 2 - 1 2 )^2+ (0- 1 2 )^2 = 1 2 + 1 2 =1

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0 and let g(x)= a x^3 3 + b x^2 2 +c x . Statement-I : The given quadratic equation ax ^2+ bx + c =0 has at least 2025 The difference between the absolute maximum and absolute minimum values of the function f(x)=2 x^3-15 x^2+36 x-30 on [-1,4] is 2025 If f(x)=x e^ x(1-x) , x R , then f(x) is 2025 The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025 If the tangent of the curve 4 y^3=3 a x^2+x^3 drawn at the point (a, a) forms a triangle of area 25 24 sq.units with the coordinate axes then a = 2025 If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025 If the Lagrange's mean value theorem is applied to the function f(x)=e^x defined on the interval [1,2] and the value of c (1,2) is k , then e^ k-1 = 2025 If the tangent to the curve x y+a x+b y=0 at (1,1) makes an angle Tan ⁻¹ 2 with X -axis, then ab a + b = 2025 Full Application of Derivatives list All AP EAMCET PYQs