Answer:
Step-by-step explanation:
(a)
[tex]y = kx^{2} -8x-5[/tex]
Derivative of the curve is:
[tex]\frac{dy}{dx} = 2kx - 8[/tex]
At point R, value of the derivative would be:
[tex]\frac{dy}{dx}=4k-8[/tex]
The derivative of a curve at a point is the slope of the tangent to the curve at that point.
∴Slope of tangent at R = 4k - 8
Now, equation of normal to the curve is: 4y + x = 24;
Convert this in y = mx + c form:
[tex]y = \frac{-x}{4} + 6[/tex]
Hence, slope of the normal is: (-1/4)
Now, relation between slope of normal and tangent is:
Slope of tangent = -1/Slope of normal
Substituting the slopes for normal and tangent in this equation we get:
[tex]4k - 8 = \frac{-1}{(-1/4)}[/tex]
Calculating, k = 3.
(b)
Point R has the x co-ordinate of 2. Since point R lies on the curve with the given equation, it must satisfy the equation.
Hence, substituting the x co-ordinate of R in the equation we get,
[tex]y=4*3-16-5=-9[/tex]
∴R(x,y) ≡ (2, -9)
Now, to find the co-ordinates of point S, the equations of the tangent to the first curve must be known. Let's assume that the equation of tangent is given by:
y = mx + c
Here, m = slope of tangent = 4k - 8 = 4;
Also, this tangent passes through point R, which has x co-ordinate 2. Hence, R must satisfy the equation of the tangent. Hence, substituting co-ordinates of R:
-9 = 4*2 + c
∴ c = -17
Hence, equation of the tangent is:
y = 4x - 17 ;
Now, equating the equation of the tangent and the 2nd curve to find the value of x co -ordinate of S:
[tex]4x - 17 = 4x -\frac{1}{x^3} - 9[/tex]
Solving for x: x = 1/2;
Substitute x = 1/2 in eq. of tangent to find the y co-ordinate:
y = 4*(1/2) - 17 = -15
S(x,y) ≡ (1/2,-15)
