Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The equation of the tangent to the curve 
y=kx+(6)/(x) at the point 
(-2,-19) is 
px+qy=c. Find the values of the integers 
k,p,q and 
c, where 
c > 0.

55. The equation of the tangent to the curve y=kx+6x y=k x+\frac{6}{x} at the point (2,19) (-2,-19) is px+qy=c p x+q y=c . Find the values of the integers k,p,q k, p, q and c c , where c>0 c>0 .

Full solution

Q. 55. The equation of the tangent to the curve y=kx+6x y=k x+\frac{6}{x} at the point (2,19) (-2,-19) is px+qy=c p x+q y=c . Find the values of the integers k,p,q k, p, q and c c , where c>0 c>0 .
  1. Find Derivative: First, we need to find the derivative of y=kx+6xy=kx+\frac{6}{x} to get the slope of the tangent at the point (2,19)(-2,-19).\newlineDifferentiate yy with respect to xx: y=k6x2y' = k - \frac{6}{x^2}.
  2. Calculate Slope: Now, plug in the xx-coordinate of the point (2)(-2) into the derivative to find the slope of the tangent.y(2)=k6(2)2=k64=k1.5.y'(-2) = k - \frac{6}{(-2)^2} = k - \frac{6}{4} = k - 1.5.
  3. Point-Slope Form: Since the tangent passes through (2,19)(-2,-19), we can use the point-slope form of the equation of a line: yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point.\newlineSubstitute mm with k1.5k - 1.5 and (x1,y1)(x_1, y_1) with (2,19)(-2, -19): y(19)=(k1.5)(x(2))y - (-19) = (k - 1.5)(x - (-2)).
  4. Simplify Equation: Simplify the equation: y+19=(k1.5)(x+2)y + 19 = (k - 1.5)(x + 2).\newlineExpand the right side: y+19=kx+2k1.5x3y + 19 = kx + 2k - 1.5x - 3.
  5. Rearrange Equation: We need to rearrange the equation to the form px+qy=cpx+qy=c. Combine like terms: (k1.5)x+y=192k+3(k - 1.5)x + y = -19 - 2k + 3.
  6. Compare with Standard Form: Now, compare the equation with px+qy=cpx+qy=c to find pp, qq, and cc.p=k1.5p = k - 1.5, q=1q = 1, and c=192k+3c = -19 - 2k + 3.
  7. Find kk: We know the point (2,19)(-2,-19) lies on the curve y=kx+6xy=kx+\frac{6}{x}, so we can plug it in to find kk.19=k(2)+62-19 = k(-2) + \frac{6}{-2}.
  8. Solve for k: Solve for kk: 19=2k3-19 = -2k - 3. Add 2k2k to both sides: 2k19=32k - 19 = -3. Add 1919 to both sides: 2k=162k = 16. Divide by 22: k=8k = 8.
  9. Find p,q,cp, q, c: Now that we have kk, we can find p,q,p, q, and cc.p=k1.5=81.5=6.5p = k - 1.5 = 8 - 1.5 = 6.5, but pp must be an integer, so there's a mistake.

More problems from Find equations of tangent lines using limits