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Question: The slope of tangent to the curve \(x = {t^2} + 3t - 8\) and \(y = 2{t^2} - 2t - 5\) at the point \(...

The slope of tangent to the curve x=t2+3t−8x = {t^2} + 3t - 8 and y=2t2−2t−5y = 2{t^2} - 2t - 5 at the point (2,−1)\left( {2, - 1} \right) is:
(A) 227\dfrac{{22}}{7}
(B) 67\dfrac{6}{7}
(C) −6 - 6
(D) None of these

Explanation

Solution

In the question, we are provided with the parametric equation of a curve and we have to find the equation of tangent at the point given to us. So, we first find the parameter with the help of coordinates of the point given to us. Then, we differentiate the expressions of x and y to find dydx\dfrac{{dy}}{{dx}}. Then, we find the slope of the tangent by putting in the value of parameter t in the expression for slope.

Complete step by step answer:
Now, we have x=t2+3t−8x = {t^2} + 3t - 8 and y=2t2−2t−5y = 2{t^2} - 2t - 5. We are given the coordinates of the point lying on the curve as (2,−1)\left( {2, - 1} \right).
So, x=t2+3t−8=2x = {t^2} + 3t - 8 = 2
Shifting all the terms to left side of the equation, we get,
⇒t2+3t−8−2=0\Rightarrow {t^2} + 3t - 8 - 2 = 0
⇒t2+3t−10=0\Rightarrow {t^2} + 3t - 10 = 0
For factorising the quadratic equation, we use the splitting the middle term method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and product of the terms gives us the product of the constant term and coefficient of x2{x^2}.
⇒t2+(5−2)t−10=0\Rightarrow {t^2} + \left( {5 - 2} \right)t - 10 = 0
We split the middle term 3t3t into two terms 5t5t and −2t - 2t since the product of these terms, −10t2 - 10{t^2} is equal to the product of the constant term and coefficient of x2{x^2} and sum of these terms gives us the original middle term, 3t3t.
⇒t2+5t−2t−10=0\Rightarrow {t^2} + 5t - 2t - 10 = 0
Taking x common from the first two terms and 77 common from the last two terms. We get,
⇒t(t+5)−2(t+5)=0\Rightarrow t\left( {t + 5} \right) - 2\left( {t + 5} \right) = 0
⇒(t+5)(t−2)=0\Rightarrow \left( {t + 5} \right)\left( {t - 2} \right) = 0
Either t+5=0t + 5 = 0 or t−2=0t - 2 = 0
⇒t=−5\Rightarrow t = - 5 or t=2t = 2
So, the value of x coordinate is 22 for t=−5t = - 5 or t=2t = 2.
Now, y=2t2−2t−5=−1y = 2{t^2} - 2t - 5 = - 1
Shifting all the terms to left side of the equation, we get,
⇒2t2−2t−5+1=0\Rightarrow 2{t^2} - 2t - 5 + 1 = 0
⇒2t2−2t−4=0\Rightarrow 2{t^2} - 2t - 4 = 0
Dividing both sides by 22, we get,
⇒t2−t−2=0\Rightarrow {t^2} - t - 2 = 0
Using splitting the middle term method, we get,
⇒t2+(1−2)t−2=0\Rightarrow {t^2} + \left( {1 - 2} \right)t - 2 = 0
We split the middle term −t - t into two terms tt and −2t - 2t since the product of these terms, −2t2 - 2{t^2} is equal to the product of the constant term and coefficient of x2{x^2} and sum of these terms gives us the original middle term, −t - t.
⇒t2+t−2t−2=0\Rightarrow {t^2} + t - 2t - 2 = 0
⇒t(t+1)−2(t+1)=0\Rightarrow t\left( {t + 1} \right) - 2\left( {t + 1} \right) = 0
⇒(t+1)(t−2)=0\Rightarrow \left( {t + 1} \right)\left( {t - 2} \right) = 0
Either t+1=0t + 1 = 0 or t−2=0t - 2 = 0
⇒t=−1\Rightarrow t = - 1 or t=2t = 2
So, the value of y coordinate is −1 - 1 for t=−1t = - 1 or t=2t = 2.
Now, since the x and y coordinates are equal to 22 and (−1)\left( { - 1} \right) simultaneously. Hence, the value of t will be equal to 22 since it is the common solution of both the equations.
Now, we differentiate the equations x=t2+3t−8x = {t^2} + 3t - 8 and y=2t2−2t−5y = 2{t^2} - 2t - 5 with respect to t.
So, we get, dxdt=ddt[t2+3t−8]\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left[ {{t^2} + 3t - 8} \right]
We know the power rule of differentiation ddx(xn)=nx(n−1)\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}}. So, we get,
⇒dxdt=2t+3−−−−(1)\Rightarrow \dfrac{{dx}}{{dt}} = 2t + 3 - - - - \left( 1 \right)
Also, y=2t2−2t−5y = 2{t^2} - 2t - 5
We know that the derivative of a constant is zero. So, we get,
⇒dydt=4t−2−−−−(2)\Rightarrow \dfrac{{dy}}{{dt}} = 4t - 2 - - - - \left( 2 \right)
Now, we divide the equation (2)\left( 2 \right) by the equation (1)\left( 1 \right).
dydx=4t−22t+3\dfrac{{dy}}{{dx}} = \dfrac{{4t - 2}}{{2t + 3}}
So, the slope of the tangent at the point can be calculated by substituting the value of parameter t into the expression dydx\dfrac{{dy}}{{dx}}. So, we get,
⇒dydx=4(2)−22(2)+3\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( 2 \right) - 2}}{{2\left( 2 \right) + 3}}
Simplifying the expression, we get,
⇒dydx=8−24+3\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{8 - 2}}{{4 + 3}}
⇒dydx=67\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{6}{7}
Therefore, the slope of tangent to the curve x=t2+3t−8x = {t^2} + 3t - 8 and y=2t2−2t−5y = 2{t^2} - 2t - 5 at the point (2,−1)\left( {2, - 1} \right) is 67\dfrac{6}{7}. Hence, option (B) is the correct answer.

Note:
We must know the procedure for calculating the slope of the tangent for the parametric form of a curve. Splitting of the middle term can be a tedious process at times when the product of the constant term and coefficient of x2{x^2} is a large number with a large number of divisors. Special care should be taken in such cases. We must know the rules of differentiation to solve the problem.