JEE Main 12 Jan 2019 Evening Question 7
Question: The tangent to the curve $ y=x^{2}-5x+5, $ parallel to the line $ 2y=4x+1, $ also passes through the point
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ ( \frac{7}{2},\frac{1}{4} ) $
B) $ ( \frac{1}{4},\frac{7}{2} ) $
C) $ ( -\frac{1}{8},7 ) $
D) $ ( \frac{1}{8},-7 ) $
Show Answer
Answer:
Correct Answer: D
Solution:
- Here, curve is $ y=x^{2}-5x+5 $
$ \Rightarrow $ $ \frac{dy}{dx}=2x-5 $
Since, the tangent is parallel to the line $ 2y=4x+1 $
$ \therefore $ $ \frac{dy}{dx}=2x-5=2 $
$ \Rightarrow $ $ x=\frac{7}{2} $
When $ x=\frac{7}{2},y={{( \frac{7}{2} )}^{2}}-5\times \frac{7}{2}+5=\frac{-1}{4} $
$ \therefore $ Equation of tangent at $ ( \frac{7}{2},\frac{-1}{4} ) $ is $ ( y+\frac{1}{4} )=2( x-\frac{7}{2} ) $
$ \Rightarrow $ $ y-2x+\frac{29}{4}=0 $
Only the point in option i.e., $ ( \frac{1}{8},-7 ) $ satisfies the above equation.
