Q110 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

The equation of the tangent to the curve $x^2+y-7=4 x$ at the point $(1,10)$ is

Correct answer
A
y = 2x + 8
B
y = x + 8
C
y = –2x – 14
D
y = x – 4

Solution

Given curve is $x^2+y-7=4 x$ now tangent at $(1,10)$ is $\begin{aligned} & 1 \mathrm{x}+\left(\frac{\mathrm{y}+10}{2}\right)-7-4\left(\frac{\mathrm{x}+1}{2}\right)=0 \\ & \Rightarrow \mathrm{y}=2 \mathrm{x}+8 \end{aligned}$

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

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