Application Of Derivatives Ques 19
19. If the tangent to the curve, $y=x^{3}+a x-b$ at the point $(1,-5)$ is perpendicular to the line, $-x+y+4=0$, then which one of the following points lies on the curve?
(a) $(-2,2)$
(b) $(2,-2)$
(c) $(-2,1)$
(d) $(2,-1)$
(2019 Main, 9 April I)
Show Answer
Answer:
Correct Answer: 19.(b)
Solution:
Formula:
Equation of tangent and normal :
- Given curve is $y=x^{3}+a x-b ……(i)$
passes through point $P(1,-5)$.
$ \therefore \quad-5=1+a-b \Rightarrow b-a=6 ……(ii) $
and slope of tangent at point $P(1,-5)$ to the curve (i), is
$ m_ {1}=\left.\frac{d y}{d x}\right|_ {(1,-5)}=\left[3 x^{2}+a\right]_ {(1,-5)}=a+3 $
$\because$ The tangent having slope $m_{1}=a+3$ at point $P(1,-5)$ is perpendicular to line $-x+y+4=0$, whose slope is $m_{2}=1$.
$\therefore \quad a+3=-1 \Rightarrow a=-4 \quad\left[\because m_{1} m_{2}=-1\right]$
Now, on substituting $a=-4$ in Eq. (ii), we get $b=2$
On putting $a=-2$ and $b=2$ in Eq. (i), we get
$ y=x^{3}-4 x-2 $
Now, from option $(2,-2)$ is the required point which lie on it.
BETA
