If the line joining the points (0,3) and (5,-2) is a tangent to the curve y=(c/x+1), then find the value of c.

Question

If the line joining the points (0,3) and (5,-2) is a tangent to the curve y=(c/x+1), then find the value of c. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

line joining (0,3)(5,-2) y-3=(-2-3/5-0)(x-0) y-3=-x y=3-x touches y =( c / x +1) (3-x)=(c/x+1) 3 x+3-x2-x=c x2-2 x+(c-3)=0 D=0 (-2)2-4 × 1(c-3)=0 4-4 c+12=0 c=4

Q. If the line joining the points $(0, 3)$ and $(5, - 2)$ is a tangent to the curve $y = \frac{c}{x + 1}$ , then find the value of $c$ .

line joining $(0, 3) (5, - 2)$
$y - 3 = \frac{- 2 - 3}{5 - 0} (x - 0)$
$y - 3 = - x$
$y = 3 - x$
touches $y = \frac{c}{x + 1}$
$(3 - x) = \frac{c}{x + 1}$
$3 x + 3 - x^{2} - x = c$
$x^{2} - 2 x + (c - 3) = 0$
$D = 0$
$(- 2)^{2} - 4 \times 1 (c - 3) = 0$
$4 - 4 c + 12 = 0$
$c = 4$

Q. If the line joining the points (0,3) and (5,−2) is a tangent to the curve y=x+1c​, then find the value of c.

line joining (0,3)(5,−2) y−3=5−0−2−3​(x−0) y−3=−x y=3−x touches y=x+1c​ (3−x)=x+1c​ 3x+3−x2−x=c x2−2x+(c−3)=0 D=0 (−2)2−4×1(c−3)=0 4−4c+12=0 c=4

Q. If the line joining the points $(0, 3)$ and $(5, - 2)$ is a tangent to the curve $y = \frac{c}{x + 1}$ , then find the value of $c$ .

line joining $(0, 3) (5, - 2)$
$y - 3 = \frac{- 2 - 3}{5 - 0} (x - 0)$
$y - 3 = - x$
$y = 3 - x$
touches $y = \frac{c}{x + 1}$
$(3 - x) = \frac{c}{x + 1}$
$3 x + 3 - x^{2} - x = c$
$x^{2} - 2 x + (c - 3) = 0$
$D = 0$
$(- 2)^{2} - 4 \times 1 (c - 3) = 0$
$4 - 4 c + 12 = 0$
$c = 4$