Question

In the curve $x={{t}^{2}}+3t-8,y=2{{t}^{2}}-2t-5$, at point (2, -1),
A. Length of sub tangent is $\dfrac{7}{6}$
B. Slope of tangent $=\dfrac{6}{7}$
C. Length of tangent $=\dfrac{\sqrt{\left( 85 \right)}}{6}$
D. Slope of tangent $=\dfrac{22}{7}$

Answer 1

Hint: Here we can see that $\left( x,y \right)$ are given in terms of another variable t. Which is making it dependent on both x and y, and point (2, -1) is given on the curve of $\left( x,y \right)$. So, first find out that common t which will validate that point on the $\left( x,y \right)$ curve, then proceed for the tangent in which first we have to find the slope, then the equation of that tangent.

Complete step by step answer:
So, as required t for the point (2, -1).
Go, with the given equations $x={{t}^{2}}+3t-8,y=2{{t}^{2}}-2t-5$.
Put $\left( x,y \right)=\left( 2,-1 \right)$ in the equations,
For the x = 2,
$2={{t}^{2}}+3t-8........\left( 1 \right)$
Root of quadratic equation (1) using formula
$\begin{align}
  & t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & t=\dfrac{-3\pm \sqrt{9-4\times 1\times \left( -10 \right)}}{2\times 1} \\
 & t=\dfrac{-3\pm \sqrt{49}}{2}=\dfrac{-3\pm 7}{2} \\
 & t=2,-5 \\
\end{align}$
Same for y = -1,
$-1=2{{t}^{2}}-2t-5.........\left( 2 \right)$
Root of quadratic equation (2) using formula
$\begin{align}
  & t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & t=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 2\times \left( -4 \right)}}{2\times 2} \\
 & t=\dfrac{2\pm \sqrt{36}}{4}=\dfrac{2\pm 6}{4} \\
 & t=2,-1 \\
\end{align}$
Now, we can see t = 2 is common for x = 2 and y = -1.
Now we have to find the slope of the tangent on point (2, -1).
$Slope=\dfrac{dy}{dx}$ for any tangent on the curve.
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\]
And we know $\dfrac{dy}{dt}=4t-2\ and\ \dfrac{dx}{dt}=2t+3$.
For the t = 2,
$\dfrac{dy}{dx}=\dfrac{4t-2\ }{2t+3}=\dfrac{4\times 2-2}{2\times 2+3}=\dfrac{6}{7}$
Now, the equation of tangent on (2, -1)
$\begin{align}
  & \left( y-\left( -1 \right) \right)=\dfrac{6}{7}\left( x-2 \right) \\
 & \left( y+1 \right)=\dfrac{6}{7}\left( x-2 \right)...............\left( 3 \right) \\
\end{align}$
Now, using equation (3) , find the intersection point on x – axis.
So, put y = 0 in equation (3)
$\begin{align}
  & \left( 0+1 \right)=\dfrac{6}{7}\left( x-2 \right) \\
 & x=\dfrac{19}{6} \\
\end{align}$
The Intersection point is $\left( \dfrac{19}{6},0 \right)$ .
So, length of tangent is equal to distance between $\left( \dfrac{19}{6},0 \right)$ and (2, -1).
Length of tangent $=\sqrt{{{\left( \dfrac{19}{6}-2 \right)}^{2}}+{{\left( 0-\left( -1 \right) \right)}^{2}}}$
$=\sqrt{\dfrac{85}{36}}=\dfrac{\sqrt{85}}{6}$
And the length of subtangent is the projection of tangent on the x – axis.
Length of Subtangent $=\left( \dfrac{19}{6}-2 \right)$
Length of Sub-tangent $=\dfrac{7}{6}$
Now, check with the given options, and we can say option (a), (b) and (c) are correct answers.

Note: In this problem we can easily go with a graphical method. When we know the slope and point of the tangent, the graphical method is easy to understand and visualise. But in this problem we can go with formula and basic knowledge to save time.