Question

The line 3y+x ,25 is a normal to the curve y=x*x-5x+k find the value of constant K

Answer 1

Answer:

Step-by-step explanation:

$\sf{Given\,\,\,that\,\,\,3\,y+x=25\,\,\,is\,\,\,a\,\,\,normal\,\,\,to\,\,\,the\,\,\,curve\,\,\,y=x^2-5x+k}$

So,

$\tt{3y=-x+25}$

$\tt{\implies\,y=\left(-\dfrac{1}{3}\right)x+\dfrac{25}{3}}$

So,

$\mapsto\,\bold{\sf{Slope\,\,of\,\,normal=\,-\dfrac{1}{3}}}$

Also, differentiating the curve w.r.t x

$\tt{\dfrac{dy}{dx}=2x-5}$

$\mapsto\,\bold{\sf{Slope\,\,of\,\,normal=\,-\dfrac{1}{2x-5}}}$

Now,

$\sf{-\dfrac{1}{3}=-\dfrac{1}{2x-5}}$

$\sf{\implies\dfrac{1}{3}=\dfrac{1}{2x-5}}$

$\sf{\implies2x-5=3}$

$\sf{\implies2x=3+5}$

$\sf{\implies2x=8}$

$\sf{\implies\,x=4}$

Put this value in the equation of normal

$\sf{3y+4=25}$

$\sf{\implies3y=25-4}$

$\sf{\implies3y=21}$

$\sf{\implies\,y=7}$

So, the point (4, 7) lies on the curve

$\tt{y=x^2-5x+k}$

$\tt{\implies\,(7)=(4)^2-5(4)+k}$

$\tt{\implies\,7=16-20+k}$

$\tt{\implies\,7=-4+k}$

$\tt{\implies\,7+4=k}$

$\tt{\implies\,k=11}$