Q48 / 90 · JEE Main 2019 (12 Jan Shift 2) · Mathematics

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 :

Solution

Equation of line parallel to given line can be taken as

$y = 2 x + c$ .

Now, this line touches the curve $y = x^{2} - 5 x + 5$ hence, roots of the equation

$x^{2} - 5 x + 5 = 2 x + c \Rightarrow x^{2} - 7 x + (5 - c) = 0$ must be equal.

$\Rightarrow D = 0 \Rightarrow {(- 7)}^{2} - 4 \cdot 1 \cdot (5 - c) = 0 \Rightarrow 4 c = - 29$

Hence, equation of tangent is $4 y = 8 x - 29$ .

Clearly, it passes through $(\frac{1}{8}, - 7)$ .

Hence, the correct answer is option: 'D'.