Application of Derivatives 1 Question 4
####4. The tangent to the curve $y=x^{2}-5 x+5$, parallel to the line $2 y=4 x+1$, also passes through the point
(a) $\frac{1}{4}, \frac{7}{2}$
(b) $\frac{7}{2}, \frac{1}{4}$
(c) $-\frac{1}{8}, 7$
(d) $\frac{1}{8},-7$
(2019 Main, 12 Jan II)
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Answer:
Correct Answer: 4. (b)
Solution:
- The given curve is $y=x^{2}-5 x+5$
Now, slope of tangent at any point $(x, y)$ on the curve is
$ \frac{d y}{d x}=2 x-5 $
[on differentiating Eq. (i) w.r.t. $x$ ]
$\because$ It is given that tangent is parallel to line
$ 2 y=4 x+1 $
So, $\quad \frac{d y}{d x}=2[\because$ slope of line $2 y=4 x+1$ is 2$]$
$\Rightarrow \quad 2 x-5=2 \Rightarrow 2 x=7 \Rightarrow x=\frac{7}{2}$
On putting $x=\frac{7}{2}$ in Eq. (i), we get
$ y=\frac{49}{4}-\frac{35}{2}+5=\frac{69}{4}-\frac{35}{2}=-\frac{1}{4} $
Now, equation of tangent to the curve (i) at point
$\frac{7}{2},-\frac{1}{4}$ and having slope 2 , is
$ \begin{aligned} y+\frac{1}{4} & =2 \quad x-\frac{7}{2} \quad \Rightarrow \quad y+\frac{1}{4}=2 x-7 \\ \Rightarrow \quad y & =2 x-\frac{29}{4} \end{aligned} $
On checking all the options, we get the point $\frac{1}{8},-7$ satisfy the line (iii).
