Hey, the question is as follows:
To the curve [math]y=2-(x-2)^2[/math] both a tangent and a normal is drawn from the same point. The tangent and the normal together shape a triangle with the area of 4.5 area units. Find the point of tangent.
I initially solved it by first expressing a function for both the tangent and the normal as follows:
[math]y_t-f(a)=f'(a)(x-a)[/math][math]y_n-f(a)=-\frac{1}{f'(a)}(x-a)[/math]. Plugging in and simplifying I get that;
[math]y_t(x) =a^2+4x-2ax-2[/math][math]y_n(x) =\frac{8-21a+12a^2-2a^3+x}{2(a-2)}[/math]
If I then find the point at which they intersect the x-axis and take the difference, I have got my base and the height is given by f(a). So a can then be given using area of a triangle.
[math]y_t=0 \implies x_t=\frac{a^2-2}{2(a-2)}[/math][math]y_n=0 \implies x_n=-8+21a-12a^2+2a^3[/math]
The area is given by [math]\frac{bh}{2}=\frac{\mid x_t-x_n \mid f(a)}{2}=\frac{9}{2} \iff a_1=1.73499, a_2=2.26501[/math]a_1 does not yield a solution to the problem, so only a_2 is a valid value. From here it is easy to find the (x,y) coordinates and even the equation for the tangent and normal.
I was then also presented by a solution that used tan(x) and tan(90-x) due to the right angle formed at the top of the triangle (the point where the normal and tangent are drawn from).This piqued my interest, are there other methods/tricks to solving this problem?
To the curve [math]y=2-(x-2)^2[/math] both a tangent and a normal is drawn from the same point. The tangent and the normal together shape a triangle with the area of 4.5 area units. Find the point of tangent.
I initially solved it by first expressing a function for both the tangent and the normal as follows:
[math]y_t-f(a)=f'(a)(x-a)[/math][math]y_n-f(a)=-\frac{1}{f'(a)}(x-a)[/math]. Plugging in and simplifying I get that;
[math]y_t(x) =a^2+4x-2ax-2[/math][math]y_n(x) =\frac{8-21a+12a^2-2a^3+x}{2(a-2)}[/math]
If I then find the point at which they intersect the x-axis and take the difference, I have got my base and the height is given by f(a). So a can then be given using area of a triangle.
[math]y_t=0 \implies x_t=\frac{a^2-2}{2(a-2)}[/math][math]y_n=0 \implies x_n=-8+21a-12a^2+2a^3[/math]
The area is given by [math]\frac{bh}{2}=\frac{\mid x_t-x_n \mid f(a)}{2}=\frac{9}{2} \iff a_1=1.73499, a_2=2.26501[/math]a_1 does not yield a solution to the problem, so only a_2 is a valid value. From here it is easy to find the (x,y) coordinates and even the equation for the tangent and normal.
I was then also presented by a solution that used tan(x) and tan(90-x) due to the right angle formed at the top of the triangle (the point where the normal and tangent are drawn from).This piqued my interest, are there other methods/tricks to solving this problem?