Question

If tangent of point (1,2) on the curve $y=a{x}^2+bx+\frac{7}{2}$ be parallel to normal at (-2,2) on the curve
$y=x^2+6x+10,$ then

• A. $a=1$
• B. $a=-1$
• C. $b=\frac{5}{3}$
• D. $b=\frac{-5}{2}$

Answer 1

Answer: (A), (D)

The correct options are
A $a=1$
D $b=\frac{-5}{2}$

Tangent of a point $(1,2)$ on curve $y=a{x}^2+bx+\frac{7}{2}$ be parallel to normal at $(-2,2)$ on curve $y^2=x^2+6x+10$ then $a$ is:

Tangent at $(1,2)$ on $y=a{x}^2+bx+\frac{7}{2}$ is

$\frac{y}{2}+\frac{2}{2}=a\left(1\right)\left(x\right)+\frac{bx}{2}+\frac{b}{2}+\frac{7}{2}$

$y+2=2ax+bx+b+7$

$y=(2a+b)x+b+5$

$Slope=m_1=2a+b$

Equation of tangent at $(-2,2)$ on $y=x^2+6x+10$

$\frac{y+2}{2}=-2x+3(x-2)+10$

$y+2=2x+8$

$y=2x+6$

$Slope=m_2=2$

Slope of normal

$m_3=\frac{-1}{m_2}=\frac{-1}{2}$

$m_1=m_3$

$2a+b=\frac{-1}{2}$

$4a+2b+1=\mathrm{0............}\left(1\right)$

Also $y=a{x}^2+bx+\frac{7}{2}$ passes through $(1,2)$

$2a+2+3=\mathrm{0...........}\left(2\right)$

$\left(1\right)-\left(2\right)$$2a-2=0$

$a=1$

$b=\frac{-5}{2}$