Question

The slope of the tangent to the curve \(x=t^2+3t-8,y=2t^2-5\) at the point \((2, −1)\) is

• A. \(\frac{22}{7}\)
• B. \(\frac{6}{7}\)
• C. \(\frac{7}{6}\)
• D. \(\frac{-6}{7}\)

Answer 1

The Correct Option is B

The correct answer is B:\(\frac{6}{7}\)
The given curve is \(x=t^2+3t-8\) and \(y=2t^2-2t-5\).
\(∴ \frac{dx}{dt}=2t+3\) and \(\frac{dy}{dt}=4t-2\)
\(∴ \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{4t-2}{2t+3}\)
The given point is \((2, −1)\)
At \(x = 2\), we have:
\(t^2+3t-8=2\)
\(t^2+3t-10=0\)
\((t-2)(t+5)=0\)
\(t=2\, or\, t=-5\)
At \(y=-1\), we have:
\(2t^2-2t-5=-1\)
\(⇒ 2t^2-2t-4=0\)
\(⇒ 2(t^2-t-2)=0\)
\(⇒ (t-2)(t+1)=0\)
\(⇒ t=2\, or\, t=1\)
The common value of t is 2.
Hence, the slope of the tangent to the given curve at point (2, −1) is
\(\frac{dy}{dx}\bigg]_{t=2}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}\)