Find area of triangle bounded by a normal and a tangent to a curve.

Quote:

Originally Posted by Punch

The tangent and normal to the curve $y=4\sqrt{x+2}$ at the point P(7,12) cut the x-axis at points M and N respectively. Calculate the area of the triangle PMN.

My attempt is...

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

Tangent : $y-12=\frac{2(x-7)}{\sqrt{x+2}}$

I guess I need some help in solving for x...

From your dy/dx you can find the slope of the tangent at x = 7. You will also then have the slope of the normal as its negative inverse. YOu should find those as 2/3 and -3/2.

You can then have the equation of the tangent as (y-12)/(x - 7) = 3/2.
Let x = 0 to find the y intercept.

Do the same for the normal.

You should now be able to find the lengths of the line segments PM and PM, which are perpendicular to each other, so easily calculating the area of the triangle.