Question

The slope of the normal to the curve $x=t^{2}+3t-8, y=2t^{2}-2t-5$ at the point $(2,-1)$ is

• A. $\frac{6}{7}$
• B. $-\frac{6}{7}$
• C. $\frac{7}{6}$
• D. $-\frac{7}{6}$

Answer 1

Answer: (A)

$\frac{6}{7}$

Answer 2

Given :
Curves are
x = t2 + 3t - 8 …..(i)
y = 2t2 - 2t - 5 …..(ii)
At (2, -1) from (i)
t2 + 3t − 10 = 0
So, t = 2 or t = -5
From (ii)
2t2 − 2t − 4 = 0
⇒ t2 − t − 2 = 0
So, t = 2 or t = −1
Now, from both the solutions , we get t = 2
Differentiating both the equations w.r.t. t, we get
$\frac{dx}{dt}=2t+3$ ….(iii)
$\frac{dy}{dt}=4t-2$ …..(iv)
Hence,
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$=\frac{4t-2}{2t+3}$ ….. we get this from (iii) and (iv)
Now, slope of tangent to the given curve is :
$\frac{dy}{dx}=\frac{4t-2}{2t+3}$
Therefore, $|\frac{dy}{dx}|_{(2,-1)}=|\frac{4t-2}{2t+3}|_{t=2}$
$=\frac{8-2}{4+3}=\frac{6}{7}$
Hence, it is the slope of tangent to the given curve at (2, -1)
So, the correct option is (A) : $\frac{6}{7}$