Question

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

Answer 1

Given equation of curve is : \(x=t^{2}+3 t-8\) and \(y=2 t^{2}-2 t-5\)
Differentiating both the equations w.r.t. '\(t\)',
\(\Rightarrow \dfrac{d x}{d t}=2 t+3\) and \(\dfrac{d y}{d t}=4 t-2\)

\(\Rightarrow \dfrac{d y}{d x}=\dfrac{\frac{d y}{d t}}{\dfrac{d x}{d t}}=\dfrac{4 t-2}{2 t+3}\)\(\quad \ldots(1)\)

Since the curve passes through the point \((2,-1)\), we get

\(2=t^{2}+3 t-8\) and \(-1=2 t^{2}-2 t-5\)
\(\Rightarrow t^{2}+3 t-10=0\) and \(t^{2}-t-2=0\)
\(\Rightarrow \left(t-2\right)\left(t+5\right)=0\) and \(\left(t-2\right)\left(t+1\right)=0\)

\(\Rightarrow t=2,-5\) and \(t=2,-1\)
\(\Rightarrow t=2\) (common value)
From equation \((1)\), we get
Slope of tangent \(\left(\dfrac{d y}{d x}\right)_{t=2}=\dfrac{4 \times 2-2}{2 \times 2+3}=\dfrac{6}{7}\)
\(\therefore\) Slope of tangent is \(\dfrac{6}{7}\)
Hence, option (b) is correct.