Question

The point on the curve $y=x^2+4$ at which the tangent is perpendicular to the line $x+2y=3$ is

• A. $(-1,5)$
• B. $(5,1)$
• C. $(1,5)$
• D. $(1,-5)$

Answer 1

Answer: (C)

$(1,5)$

Given,

$y=x^2+4$

$\frac{dy}{dx}=2x$

let the points on the curve be, $\alpha ,\beta$

${\frac{dy}{dx}}_{\alpha ,\beta }=2\alpha =m_1$

$x+2y=3$

slope of the above line,

$m_2=-\frac{1}{2}$

we have,

$m_1{m}_2=-1$

$m_1(-\frac{1}{2})-1$

$m_1=2$

$\Rightarrow 2\alpha =2$

$\therefore \alpha =1$

$\Rightarrow \beta =(1{)}^2+4=5$

Therefore the point is $(1,5)$