Question

The slope of the normal to the curve $x = {t^2} + 3t - 8$and $y = 2{t^2} - 2t - 5$at the point $\left( {2, - 1} \right)$ is
A) $\dfrac{6}{7}$
B) $ - \dfrac{6}{7}$
C) $\dfrac{7}{6}$
D) $ - \dfrac{7}{6}$
E) $ - \dfrac{1}{2}$

Answer 1

Hint: In this question first differentiate both x and y with respect to t then combines both the derivatives to get$\dfrac{{dy}}{{dx}}$. Since, the curve passes through the point $\left( {2, - 1} \right)$ equate both the equation with the point $\left( {2, - 1} \right)$ coordinate wise to get the different value of t. Then substitute the common value of t in$\dfrac{{dy}}{{dx}}$,after that substitute the obtained value of $\dfrac{{dy}}{{dx}}$in the formula of slope of normal i.e. \[\dfrac{{ - 1}}{{\dfrac{{dy}}{{dx}}}}\].

Complete step by step solution:
Given curves are$x = {t^2} + 3t - 8$ and $y = 2{t^2} - 2t - 5$
Differentiate both the curves with respect to t.
We get,
$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {{t^2} + 3t - 8} \right)$
$ \Rightarrow \dfrac{{dx}}{{dt}} = 2t + 3$
Now on differentiating the other curve with respect to t, we get,
$\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left( {2{t^2} - 2t - 5} \right)$
$ \Rightarrow \dfrac{{dy}}{{dt}} = 4t - 2$
Slope of tangent$ = \dfrac{{dy}}{{dx}}$
$\dfrac{{dy}}{{dx}}$
On substituting the value of $\dfrac{{dy}}{{dt}}\;\text{and}\;\dfrac{{dx}}{{dt}}$, we get,
$ \Rightarrow \dfrac{{4t - 2}}{{2t + 3}} \ldots \left( 1 \right)$
Since, curve passes through the point $\left( {2, - 1} \right)$
Therefore, on factorising ${t^2} + 3t - 8 = 2$and $2{t^2} - 2t - 5 = - 1$,
Consider, ${t^2} + 3t - 8 = 2$
$ \Rightarrow {t^2} + 3t - 10 = 0$
On splitting the middle term, we get,
$ \Rightarrow $${t^2} + 5t - 2t - 10 = 0$
$ \Rightarrow \left( {t + 5} \right)\left( {t - 2} \right) = 0$
$ \Rightarrow t = - 5,2$
Now, $2{t^2} - 2t - 5 = - 1$
$ \Rightarrow 2{t^2} - 2t - 4 = 0$
Dividing by 2, we get,
$ \Rightarrow {t^2} - t - 2 = 0$
On splitting the middle term, we get,
$ \Rightarrow $${t^2} - 2t + t - 2 = 0$
$ \Rightarrow $$\left( {t + 1} \right)\left( {t - 2} \right) = 0$
$ \Rightarrow $ $t = 2, - 1$
So, common value of t is 2.
On putting value of t$ = 2$ in equation $\left( 1 \right)$ we get,
${\left[ {\dfrac{{{\text{d}}y}}{{{\text{d}}x}}} \right]_{t = 2}} = \dfrac{{4\left( 2 \right) - 2}}{{2\left( 2 \right) + 3}}$
$ = \dfrac{6}{7}$
Thus, the slope of the normal to the given curve is $ - \dfrac{7}{6}$.
Hence, Option D $ - \dfrac{7}{6}$ is a correct answer.

Note: The normal line is defined as the line that is perpendicular to the tangent line at the point of tangency. Because the slopes of perpendicular lines are negative reciprocals of one another. Therefore the slope of normal is negative reciprocal of the slope of the tangent.