Question

The two curves $y=x^2-1$ and $y=8x-x^2-9$ at the point $(2,3)$ have common

• A. tangent as $4x-y-5=0$
• B. tangent as $x+4y-14=0$
• C. normal as $4x+y=11$
• D. normal as $x-4y=10$

Answer 1

Answer: (A)

tangent as $4x-y-5=0$

$y_1=x^2-1$

$\therefore y_1^{\prime }=2x$

$y_2=8x-x^2-9$

$\therefore y_2^{\prime }=8-2x$

At the common tangent point, slope will be same.
$2x=8-2x$
$4x=8$
$x=2,\therefore y=3$
This point, $(2,3)$, satisfies both curves.
At $(2,3)$
$y_1^{\prime }=4$
Equation of tangent at $(2,3)$
$(y-3)=4(x-2)$
$4x-y-5=0$

For normal, slope will be $-\frac{1}{4}$
$\therefore$ Equation of normal at $(2,3)$ is
$(y-3)=\frac{-1}{4}(x-2)$
$4y-12+x-2=0$
$4y+x-14=0$

Hence, option A.