Find the equation of the tangent to the curve x2 - xy - y2 + 4 = 0 at point K(2,4).

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Q1: To find the equation of the tangent to the curve x² − xy − y² + 4 = 0 at point K(2,4), we first need to find the slope of the tangent using implicit differentiation. Step 1: Differentiate the equation with respect to x. d/dx(x²) − d/dx(xy) − d/dx(y²) + d/dx(4) = d/dx(0); 2x − (1 · y + x · dy/dx) − 2y dy/dx + 0 = 0; 2x − y − x dy/dx − 2y dy/dx = 0 Step 2: Isolate dy/dx. 2x − y = (x + 2y) dy/dx; dy/dx = (2x − y)/(x + 2y) Step 3: Substitute the coordinates of point K(2,4) into the derivative to find the slope m. m = (2(2) − 4)/(2 + 2(4)) = (4 − 4)/(2 + 8) = 0/10 = 0; Step 4: Use the point-slope form of a linear equation, y − y₁ = m(x − x₁), with m=0 and (x₁, y₁) = (2,4). y − 4 = 0(x − 2); y − 4 = 0; y = 4; The equation of the tangent is y=4. Q2: We are given the sum of the first n terms of an A.P. as S_n = n² − 8n. We need to find the least possible value of $

Find the equation of the tangent to the curve x2 − xy − y2 + 4 = 0 at point K(2,4).

Answer

Step-by-step solution

Differentiate the equation with respect to $x$ .

Isolate $\frac{dy}{dx}$ .

Substitute the coordinates of point $K(2,4)$ into the derivative to find the slope $m$ .

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , with $m=0$ and $(x_1, y_1) = (2,4)$ .

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